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Matric S

The document provides an overview of matrix operations, including addition, subtraction, scalar multiplication, and multiplication of matrices. It defines key concepts such as equality of matrices and introduces applications of matrices in various fields. Additionally, it includes examples to illustrate the practical use of matrices in solving systems of equations and organizing information.

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Matric S

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2 Matrices

2.1 Operations with Matrices

2.2 Properties of Matrix Operations
2.3 The Inverse of a Matrix
2.4 Elementary Matrices
2.5 Markov Chains
2.6 More Applications of Matrix Operations

Data Encryption (p. 94)

Computational Fluid Dynamics (p. 79)

Beam Deflection (p. 64)

Information Retrieval (p. 58)

Flight Crew Scheduling (p. 47)

Clockwise from top left, Cousin_Avi/Shutterstock.com; Goncharuk/Shutterstock.com;
39
Gunnar Pippel/Shutterstock.com; Andresr/Shutterstock.com; nostal6ie/Shutterstock.com

Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
40 Chapter 2 Matrices

2.1 operations with Matrices

Determine whether two matrices are equal.
Add and subtract matrices and multiply a matrix by a scalar.
Multiply two matrices.
Use matrices to solve a system of linear equations.
Partition a matrix and write a linear combination of column vectors.

Equality of MatricES
In Section 1.2, you used matrices to solve systems of linear equations. This chapter
introduces some fundamentals of matrix theory and further applications of matrices.
It is standard mathematical convention to represent matrices in any one of the three
ways listed below.
1. An uppercase letter such as A, B, or C
2. A representative element enclosed in brackets, such as [aij], [bij], or [cij]
3. A rectangular array of numbers

[ ]
a11 a12 . . . a1n
a21 a22 . . . a2n
⋮ ⋮ ⋮
am1 am2 . . . amn

As mentioned in Chapter 1, the matrices in this text are primarily real matrices. That
is, their entries are real numbers.
Two matrices are equal when their corresponding entries are equal.

definition of Equality of Matrices

Two matrices A = [aij] and B = [bij] are equal when they have the same size
(m × n) and aij = bij for 1 ≤ i ≤ m and 1 ≤ j ≤ n.

Equality of Matrices
rEMarK Consider the four matrices
The phrase “if and only if”
means the statement is true in
both directions. For example,
A= [13 2
4
,] B= [13], C = [1 3], and D= [1x 2
4
.]
“p if and only if q” means that
p implies q and q implies p. Matrices A and B are not equal because they are of different sizes. Similarly, B and C
are not equal. Matrices A and D are equal if and only if x = 3.

A matrix that has only one column, such as matrix B in Example 1, is a column
matrix or column vector. Similarly, a matrix that has only one row, such as matrix
C in Example 1, is a row matrix or row vector. Boldface lowercase letters often
designate column matrices and row matrices. For instance, matrix A in Example 1 can be
partitioned into the two column matrices a1 =
1
3
and a2 =
2
4
as shown below. [] []
A= [13 2
4
=] [
1
3
2
4 ]
= [a1 a2]

Matrix Addition, Subtraction,

and Scalar Multiplication
To add two matrices (of the same size), add their corresponding entries.

Definition of Matrix Addition

If A = [aij] and B = [bij] are matrices of size m × n, then their sum is the
m × n matrix A + B = [aij + bij].
The sum of two matrices of different sizes is undefined.

Addition of Matrices

a. [−10 2
1 ] [
+
1
−1
3
2 ] [
=
−1 + 1
0 + (−1)
2+3
1+2
=
0
−1 ] [ 5
3 ]
−2 −2
b. [01 1
2 3
+
0
0 ] [ 0
0
0
0 ] [
=
0
1
1
2 3 ]
REMARK

[ ] [ ] []
1 −1 0
It is often convenient to rewrite
the scalar multiple cA by
c. −3 + 3 = 0 d. [24 1
0
0
−1
+
0
−1] [ 1
3 ]
is undefined.
−2 2 0
factoring c out of every entry
in the matrix. For example,
factoring the scalar 12 out of When working with matrices, real numbers are referred to as scalars. To multiply
the matrix below gives a matrix A by a scalar c, multiply each entry in A by c.

[ ]
1
− 32 −3
2
5
2
1
2
= 12 [15 1
. ] Definition of Scalar Multiplication
If A = [aij] is an m × n matrix and c is a scalar, then the scalar multiple of A by
c is the m × n matrix cA = [caij].

You can use −A to represent the scalar product (−1)A. If A and B are of the same
size, then A − B represents the sum of A and (−1)B. That is, A − B = A + (−1)B.

Scalar Multiplication and Matrix Subtraction

For the matrices A and B, find (a) 3A, (b) −B, and (c) 3A − B.

[ ] [ ]
1 2 4 2 0 0
A = −3 0 −1 and B= 1 −4 3
2 1 2 −1 3 2

solution

[ ] [ ] [ ]
1 2 4 3(1) 3(2) 3(4) 3 6 12
a. 3A = 3 −3 0 −1 = 3(−3) 3(0) 3(−1) = −9 0 −3
2 1 2 3(2) 3(1) 3(2) 6 3 6

[ ] [ ]
2 0 0 −2 0 0
b. −B = (−1) 1 −4 3 = −1 4 −3
−1 3 2 1 −3 −2

[ ] [ ] [ ]
3 6 12 2 0 0 1 6 12
c. 3A − B = −9 0 −3 − 1 −4 3 = −10 4 −6
6 3 6 −1 3 2 7 0 4

Matrix Multiplication
Another basic matrix operation is matrix multiplication. To see the usefulness of
this operation, consider the application below, in which matrices are helpful for
organizing information.
A football stadium has three concession areas, located in the south, north, and west
stands. The top-selling items are peanuts, hot dogs, and soda. Sales for one day are
given in the first matrix below, and the prices (in dollars) of the three items are given
in the second matrix.
Numbers of Items Sold
Peanuts Hot Dogs Sodas Selling Price

[ ][ ]
South Stand 120 250 305 2.00 Peanuts
North Stand 207 140 419 3.00 Hot Dogs
West Stand 29 120 190 2.75 Soda

To calculate the total sales of the three top-selling items at the south stand, multiply
each entry in the first row of the matrix on the left by the corresponding entry in the
price column matrix on the right and add the results. The south stand sales are
(120)(2.00) + (250)(3.00) + (305)(2.75) = $1828.75 South stand sales

Similarly, the sales for the other two stands are shown below.
(207)(2.00) + (140)(3.00) + (419)(2.75) = $1986.25 North stand sales
(29)(2.00) + (120)(3.00) + (190)(2.75) = $940.50 West stand sales

The preceding computations are examples of matrix multiplication. You can write
the product of the 3 × 3 matrix indicating the number of items sold and the 3 × 1
matrix indicating the selling prices as shown below.

[ ][ ] [ ]
120 250 305 2.00 1828.75
207 140 419 3.00 = 1986.25
29 120 190 2.75 940.50
The product of these matrices is the 3 × 1 matrix giving the total sales for each of the
three stands.
The definition of the product of two matrices shown below is based on the ideas
just developed. Although at first glance this definition may seem unusual, you will see
that it has many practical applications.

Definition of Matrix Multiplication

If A = [aij] is an m × n matrix and B = [bij] is an n × p matrix, then the product
AB is an m × p matrix
AB = [cij]
where
n
cij = ∑a b
k=1
ik kj

= ai1b1j + ai2b2j + ai3b3j + . . . + ainbnj.

This definition means that to find the entry in the ith row and the jth column of
the product AB, multiply the entries in the ith row of A by the corresponding entries in
the jth column of B and then add the results. The next example illustrates this process.

finding the product of two Matrices

Find the product AB, where

[ ]
−1 3
A= 4 −2 and B= [−3
−4
2
1
.]
5 0

Solution
First, note that the product AB is defined because A has size 3 × 2 and B has size 2 × 2.
Moreover, the product AB has size 3 × 2, and will take the form

[ ][ [ ]
−1 3 c11 c12
−3
4 −2
−4
2
1 ]
= c21 c22 .
5 0 c31 c32
To find c11 (the entry in the first row and first column of the product), multiply
corresponding entries in the first row of A and the first column of B. That is,
c11 = (−1)(−3) + (3)(−4) = −9

[ ][ [ ]
−1 3 −9 c12
−3
4 −2
−4
2
1 ]
= c21 c22 .
5 0 c31 c32
Similarly, to find c12, multiply corresponding entries in the first row of A and the
second column of B to obtain
arthur cayley
c12 = (−1)(2) + (3)(1) = 1
(1821–1895)

[ ][ [ ]
British mathematician −1 3 −9 1
−3
Arthur Cayley is credited
with giving an abstract
4 −2
−4
2
1 ]
= c21 c22 .
5 0 c31 c32
definition of a matrix.
Cayley was a Cambridge Continuing this pattern produces the results shown below.
University graduate and a
lawyer by profession. He c21 = (4)(−3) + (−2)(−4) = −4
began his groundbreaking c22 = (4)(2) + (−2)(1) = 6
work on matrices as he c31 = (5)(−3) + (0)(−4) = −15
studied the theory of c32 = (5)(2) + (0)(1) = 10
transformations. Cayley
also was instrumental The product is
in the development of

[ ][ [ ]
determinants (discussed in
−1 3 −9 1
−3
Chapter 3). Cayley and two AB = 4 −2
−4
2
1
=] −4 6 .
American mathematicians, 5 0 −15 10
Benjamin Peirce
(1809–1880) and his
Be sure you understand that for the product of two matrices to be defined, the
son, Charles S. Peirce
number of columns of the first matrix must equal the number of rows of the second
(1839–1914),
are credited matrix. That is,
with developing A B = AB.
“matrix algebra.” m×n n×p m×p

Equal
Size of AB
So, the product BA is not defined for matrices such as A and B in Example 4.

Photo Researchers/Getty Images

The general pattern for matrix multiplication is shown below. To obtain the
element in the ith row and the jth column of the product AB, use the ith row of A and the
jth column of B.

[ ][ ]
a11 a12 a13 . . . a1n c11 c12 . . . c1j . . . c1p

][
b11 b12 . . . b1j . . . b1p
a21 a22 a23 . . . a2n c21 c22 . . . c2j . . . c2p
b21 b22 . . . b2j . . . b2p
⋮ ⋮ ⋮ ⋮ b31 b32 . . . b3j . . . b3p = ⋮ ⋮ ⋮ ⋮
ai1 ai2 ai3 . . . ain ci1 ci2 . . . cij . . . cip
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
bn1 bn2 . . . bnj . . . bnp
am1 am2 am3 . . . amn cm1 cm2 . . . cmj . . . cmp

ai1b1j + ai2b2j + ai3b3j + . . . + ainbnj = cij

DIS CO VERY
Let

A= [13 2
4 ] and B = [01 1
2 ]
.

1. Find A + B and B + A. Is matrix addition commutative?

2. Find AB and BA. Is matrix multiplication commutative?

Matrix Multiplication

See LarsonLinearAlgebra.com for an interactive version of this type of example.

[ ]
−2 4 2
−5 −1
a. [ 1
2
0
−1
3
−2 ] 1 0 0 =
−3 [ 7
6 6 ]
−1 1 −1
2×3 3×3 2×3

b. [−23 4
5 ][10 0
1 ] [
=
3
−2 ]
4
5
2×2 2×2 2×2

c. [11 2
1 ][−11 2
−1
=
1
0] [ 0
1 ]
2×2 2×2 2×2

[ ]
2
d. [1 −2 −3] −1 = [1]
1
1×3 3×1 1×1

[ ] [ ]
2 2 −4 −6
e. −1 [1 −2 −3] = −1 2 3
1 1 −2 −3
3×1 1×3 3×3

Note the difference between the two products in parts (d) and (e) of Example 5.
In general, matrix multiplication is not commutative. It is usually not true that the
product AB is equal to the product BA. (See Section 2.2 for further discussion of
the noncommutativity of matrix multiplication.)

Systems of Linear Equations

One practical application of matrix multiplication is representing a system of linear
equations. Note how the system
a11x1 + a12x2 + a13x3 = b1
a21x1 + a22x2 + a23x3 = b2
a31x1 + a32x2 + a33x3 = b3
can be written as the matrix equation Ax = b, where A is the coefficient matrix of the
system, and x and b are column matrices.

[ ][ ] [ ]
a11 a12 a13 x1 b1
a21 a22 a23 x2 = b2
a31 a32 a33 x3 b3
A x = b

Solving a System of Linear Equations

Solve the matrix equation Ax = 0, where

[]
x1
−2
A= [ 1
2 3
1
−2 ]
, x = x2 , and 0 =
0
0
. []
x3
solution
As a system of linear equations, Ax = 0 is
x1 − 2x2 + x3 = 0
2x1 + 3x2 − 2x3 = 0.
Using Gauss-Jordan elimination on the augmented matrix of this system, you obtain
technology
Many graphing utilities and
software programs can
perform matrix addition, scalar
[ 1
0
0
1
− 17
− 47
0
0 ]
.

multiplication, and matrix So, the system has infinitely many solutions. Here a convenient choice of a parameter
multiplication. When you use is x3 = 7t, and you can write the solution set as
a graphing utility to check one
of the solutions in Example 6,
x1 = t, x2 = 4t, x3 = 7t, t is any real number.
you may see something similar In matrix terminology, you have found that the matrix equation
to the screen below.

[]
x
−2 1 1
[A]
[[1 -2 1 ]
[ 1
2 3
x =
−2 2 ]
0
0 []
[2 3 -2]] x3
[B]
[[1] has infinitely many solutions represented by
[4]
[7]]

[ ] [ ] []
[A]*[B] x1 t 1
[[0] x = x2 = 4t = t 4 , t is any scalar.
[0]]
x3 7t 7
The Technology Guide at That is, any scalar multiple of the column matrix on the right is a solution. Here are
CengageBrain.com can help some sample solutions:
you use technology to perform

[] [ ] [] [ ]
matrix operations. 1 2 0 −1
4 , 8 , 0 , and −4 .
7 14 0 −7

Partitioned Matrices
The system Ax = b can be represented in a more convenient way by partitioning the
matrices A and x in the manner shown below. If

[ ] [] []
a11 a12 . . . a1n x1 b1
a21 a22 . . . a2n x2 b2
A= , x= , and b =
⋮ ⋮ ⋮ ⋮ ⋮
am1 am2 . . . amn xn bm
are the coefficient matrix, the column matrix of unknowns, and the right-hand side,
respectively, of the m × n linear system Ax = b, then

[ ][ ]
a11 a12 . . . a1n x1
a21 a22 . . . a2n x2
=b
⋮ ⋮ ⋮ ⋮
am1 am2 . . . amn xn

[ ]
a11x1 + a12 x2 + . . . + a1n xn
a21x1 + a22 x2 + . . . + a2n xn
=b
⋮
am1x1 + am2 x2 + . . . + amn xn

[] [] []
a11 a12 a1n
a21 a22 a2n
x1 + x2 + . . . + xn = b.
⋮ ⋮ ⋮
am1 am2 amn
In other words,
Ax = x1a1 + x2a2 + . . . + xnan = b
where a1, a2, . . . , an are the columns of the matrix A. The expression

[] [] []
a11 a12 a1n
a21 a22 a2n
x1 + x2 + . . . + xn
⋮ ⋮ ⋮
am1 am2 amn
is called a linear combination of the column matrices a1, a2, . . . , an with coefficients
x1, x2, . . . , xn.

Linear Combinations of Column Vectors

The matrix product Ax is a linear combination of the column vectors a1,
a2, . . . , an that form the coefficient matrix A.

[] [] []
a11 a12 a1n
a21 a22 a2n
x1 + x2 + . . . + xn
⋮ ⋮ ⋮
am1 am2 amn
Furthermore, the system
Ax = b
is consistent if and only if b can be expressed as such a linear combination, where
the coefficients of the linear combination are a solution of the system.

Solving a System of linear Equations

The linear system

x1 + 2x2 + 3x3 = 0
4x1 + 5x2 + 6x3 = 3
7x1 + 8x2 + 9x3 = 6
can be rewritten as a matrix equation Ax = b, as shown below.

[] [] [] []
1 2 3 0
x1 4 + x2 5 + x3 6 = 3
7 8 9 6

Using Gaussian elimination, you can show that this system has infinitely many
solutions, one of which is x1 = 1, x2 = 1, x3 = −1.

[] [] [] []
1 2 3 0
1 4 + 1 5 + (−1) 6 = 3
7 8 9 6
That is, b can be expressed as a linear combination of the columns of A. This
representation of one column vector in terms of others is a fundamental theme of
linear algebra.

Just as you partition A into columns and x into rows, it is often useful to consider
an m × n matrix partitioned into smaller matrices. For example, you can partition the
matrix below as shown.

[ ] [ ]
1 2 0 0 1 2 0 0
3 4 0 0 3 4 0 0
−1 −2 2 1 −1 −2 2 1
You can also partition the matrix into column matrices

[ ]
1 2 0 0
3 4 0 0 = [c1 c2 c3 c4]
−1 −2 2 1
or row matrices

[ ] []
1 2 0 0 r1
3 4 0 0 = r2 .
−1 −2 2 1 r3

Many real-life applications of linear systems involve

linEar enormous numbers of equations and variables. For
algEbra example, a flight crew scheduling problem for American
appliEd Airlines required the manipulation of a matrix with
837 rows and more than 12,750,000 columns. To solve
this application of linear programming, researchers
partitioned the problem into smaller pieces and solved
it on a computer. (Source: Very Large-Scale Linear
Programming. A Case Study in Combining Interior Point and
Simplex Methods, Bixby, Robert E., et al., Operations Research,
40, no. 5)
Andresr/Shutterstock.com

2.1 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

Equality of Matrices In Exercises 1–4, find x and y. Finding Products of Two Matrices In Exercises 15–28,
find, if possible, (a) AB and (b) BA.
x −2 −4 −2
1. [ = ] [ ] −1
7 y 7 22 15. A = [14 22], B = [−12 8]
−5 x −5
2. [ = ] [ 13
] 16. A = [
2 −2
4]
, B=[
−2]
4 1
y 8 12 8
−1 2

[ ] [ ]
16 4 5 4 16 4 2x + 1 4

[ ] [ ]
2 −1 3 0 1 2
3. −3 13 15 6 = −3 13 15 3x 17. A = 5 1 −2 , B = −4 1 3
0 2 4 0 0 2 3y − 5 0 2 2 3 −4 −1 −2

[ ] [ ]
x+2 8 −3 2x + 6 8 −3

[ ] [ ]
1 −1 7 1 1 2
4. 1 2y 2x = 1 18 −8 18. A = 2 −1 8 , B= 2 1 1
7 −2 y + 2 7 −2 11 3 1 −1 1 −3 2

[ ] ] [
Operations with Matrices In Exercises 5–10, find, if 2 1 0 −1 0
possible, (a) A + B, (b) A − B, (c) 2A, (d) 2A − B, and 19. A = −3 4 , B= 4 0 2
(e) B + 12A. 1 6 8 −1 7

[−34 −2
[12 ] ]
[ ] [ ]
2 3 2 1 1 2
5. A = , B=
1 2 20. A = −3 0 4 , B= 2 −1
−2 −4 −2

[ ] [ ]
6 −1 1 4 4 1
6. A = 2 4 , B = −1 5

[]
2
−3 5 1 10 21. A = [3 2 1], B= 3
−3
7. A = [−12 1
−1
1
4
,] B= [−32 1
4
−2 ] 0

[]
−1

[ ] [ ]
3 2 −1 0 2 1 2
8. A = 2 4 5 , B= 5 4 2 22. A = , B = [2 1 3 2]
−2
0 1 2 2 1 0 1
−1
[−16 0 3
] [84 ]
[ ]
9. A = , B= −1 3
−4 0 −3 23. A = 4 −5 , B= [10 2
7 ]
[ ]
3 0 2
10. A = 2 , B = [−4 6 2]

[ ]
2 1
−1 −3
11. Find (a) c21 and (b) c13, where C = 2A − 3B,
24. A = [ 2
5 2
, ] B= 1 3
2 −1
2 −7
5
[
4 4 1
] [ ]
[ ] [ ]
A= , and B = . 0 −1 0 2
−3 1 2 0 −5 1
25. A = 4 0 2 , B = −3
12. Find (a) c23 and (b) c32, where C = 5A + 2B, 8 −1 7 1

[ ] [ ]
4 11 −9 1 0 5

[ ] [ ]
2 1 2 4 0 1 3
A= 0 3 2 , and B = −4 6 11 .
26. A = 3 −1 −2 , B = −1 2 −3 −1
−3 1 1 −6 4 9
−2 1 −2 −2 1 4 3
13. Solve for x, y, and z in the matrix equation

[]
6
4 [ x
z
y
−1
=2 ]
y
−x [ z
1
+2]4
5 [ x
−x
. ] 27. A =
−2
, B = [10 12]
1
14. Solve for x, y, z, and w in the matrix equation
6
−4
[wy x
x] [
=
2
3
−1
+2
y
z ] [ w
x
. ] 28. A = [16 0 3 −2 4
, ] B= [14 6
]
13 8 −17 20 2

Matrix Size In Exercises 29–36, let A, B, C, D, and E Writing a Linear Combination In Exercises 49–52,
be matrices with the sizes shown below. write the column matrix b as a linear combination of the
A: 3 × 4 B: 3 × 4 C: 4 × 2 D: 4 × 2 E: 4 × 3 columns of A.
−1 −1
If defined, determine the size of the matrix. If not
defined, explain why.
49. A = [13 −3
2
1 ]
, b=
7 [ ]

[ [] ]
29. A + B 30. C + E 1 2 4 1
1
31. 2D 32. −4A 50. A = −1 0 2 , b= 3
33. AC 34. BE 0 1 3 2

[ ] []
35. E − 2A 36. 2D + C 1 1 −5 3
51. A = 1 0 −1 , b = 1
Solving a Matrix Equation In Exercises 37 and 38,
2 −1 −1 0
solve the matrix equation Ax = 0.

[ ] [ ]
−3 5 −22

[]
x1
−1 −1
37. A =
2
1 [ −2 2 ]
, x = x2 , 0 =
0
0 [] 52. A = 3
4
4 , b=
−8
4
32
x3

[]
x1 Solving a Matrix Equation In Exercises 53 and 54,

[ ] []
1 2 1 3 0
x solve for A.
38. A = 1 −1 0 1 , x= 2 , 0= 0
x3
0 1 −1 2
x4
0
53. [13 ] [
2
5
A=
1
0
0
1]
Solving a System of Linear Equations In Exercises −1
54. [
−2]
A=[
1]
2 1 0
39–48, write the system of linear equations in the form 3 0
Ax = b and solve this matrix equation for x.
39. −x1 + x2 = 4 40. 2x1 + 3x2 = 5 Solving a Matrix Equation In Exercises 55 and 56,
−2x1 + x2 = 0 x1 + 4x2 = 10 solve the matrix equation for a, b, c, and d.
41. −2x1 − 3x2 = −4 42. −4x1 + 9x2 = −13
6x1 + x2 = −36 x1 − 3x2 = 12
55. [13 2
4] [ac b
d] [
=
6
19
3
2 ]
56. [
d] [ 3 1] [ 4 ]
43. x1 − 2x2 + 3x3 = 9 a b 2 1 3 17
=
c −1
−x1 + 3x2 − x3 = −6
2x1 − 5x2 + 5x3 = 17 Diagonal Matrix In Exercises 57 and 58, find the
44. x1 + x2 − 3x3 = −1 product AA for the diagonal matrix. A square matrix

[ ]
−x1 + 2x2 = 1 a11 0 0 . . . 0
x1 − x2 + x3 = 2 0 a22 0 . . . 0
45. x1 − 5x2 + 2x3 = −20 A= 0 0 a33 . . . 0
−3x1 + x2 − x3 = 8
⋮ ⋮ ⋮ ⋮
0 0 0 . . . ann
−2x2 + 5x3 = −16
is a diagonal matrix when all entries that are not on the
46. x1 − x2 + 4x3 = 17 main diagonal are zero.
x1 + 3x2 = −11

[ ] [ ]
−1 0 0 2 0 0
−6x2 + 5x3 = 40 57. A = 0 2 0 58. A = 0 −3 0
47. 2x1 − x2 + x4 = 3 0 0 3 0 0 0
3x2 − x3 − x4 = −3 Finding Products of Diagonal Matrices In Exercises
x1 + x3 − 3x4 = −4 59 and 60, find the products AB and BA for the diagonal
x1 + x2 + 2x3 = 0 matrices.
48. x1 + x2
x2 + x3
=0
=0
59. A = [20 0
−3
, ] B= [−50 0
4 ]

[ ] [ ]
x3 + x4 =0 3 0 0 −7 0 0
x4 + x5 = 0 60. A = 0 −5 0 , B= 0 4 0
0 0 0 0 0 12
−x1 + x2 − x3 + x4 − x5 = 5

61. Guided Proof Prove that if A and B are diagonal 72. Show that no 2 × 2 matrices A and B exist that satisfy
matrices (of the same size), then AB = BA. the matrix equation
Getting Started: To prove that the matrices AB and BA
are equal, you need to show that their corresponding
AB − BA =
1
0
0
1
. [ ]
entries are equal.
73. Exploration Let i = √−1 and let
(i) Begin your proof by letting A = [aij] and B = [bij]
0 −i
be two diagonal n × n matrices. A=
0
i
[
0
i
and B =
i] 0
. [ ]
(ii) The ijth entry of the product AB is
n
(a) Find A2, A3, and A4. (Note: A2 = AA,
cij = ∑a b .
k=1
ik kj
A3 = AAA = A2A, and so on.) Identify any similarities
with i2, i3, and i4.
(iii) Evaluate the entries cij for the two cases i ≠ j and (b) Find and identify B2.
i = j. 74. Guided Proof Prove that if the product AB is a square
(iv) Repeat this analysis for the product BA. matrix, then the product BA is defined.
62. Writing Let A and B be 3 × 3 matrices, where A is Getting Started: To prove that the product BA is
diagonal. defined, you need to show that the number of columns
(a)
Describe the product AB. Illustrate your answer of B equals the number of rows of A.
with examples. (i) Begin your proof by noting that the number of
(b) Describe the product BA. Illustrate your answer columns of A equals the number of rows of B.
with examples. (ii) Then assume that A has size m × n and B has size
(c) How do the results in parts (a) and (b) change when n × p.
the diagonal entries of A are all equal? (iii) Use the hypothesis that the product AB is a
square matrix.
Trace of a Matrix In Exercises 63 – 66, find the
75. Proof Prove that if both products AB and BA are
trace of the matrix. The trace of an n × n matrix A
defined, then AB and BA are square matrices.
is the sum of the main diagonal entries. That is,
Tr(A) = a11 + a22 + . . . + ann. 76. Let A and B be matrices such that the product AB is
defined. Show that if A has two identical rows, then the

[ ] [ ]
1 2 3 1 0 0 corresponding two rows of AB are also identical.
63. 0 −2 4 64. 0 1 0
77. Let A and B be n × n matrices. Show that if the ith
3 1 3 0 0 1
row of A has all zero entries, then the ith row of AB

[ ] [ ]
1 0 2 1 1 4 3 2 will have all zero entries. Give an example using 2 × 2
0 1 −1 2 4 0 6 1 matrices to show that the converse is not true.
65. 66.
4 2 1 0 3 6 2 1
0 0 5 1 2 1 1 −3
78. C
APSTONE Let matrices A and B be of
67. Proof Prove that each statement is true when A and B sizes 3 × 2 and 2 × 2, respectively. Answer each
are square matrices of order n and c is a scalar. question and explain your answers.
(a) Tr(A + B) = Tr(A) + Tr(B) (a) Is it possible that A = B?
(b) Tr(cA) = cTr(A) (b) Is A + B defined?
68. Proof Prove that if A and B are square matrices of (c) Is AB defined? If so, is it possible that AB = BA?
order n, then Tr(AB) = Tr(BA).
69. Find conditions on w, x, y, and z such that AB = BA for 79. Agriculture A fruit grower raises two crops, apples
the matrices below. and peaches. The grower ships each of these crops to
three different outlets. In the matrix
A=
w
[
y
x
z
and B =] −1
1 1
1 [ ]
70. Verify AB = BA for the matrices below.
A= [
125 100 75
100 175 125 ]
aij represents the number of units of crop i that the
α −sin α β −sin β
A= [cos
sin α cos α ] and B= [cos
sin β cos β ] grower ships to outlet j. The matrix
B = [$3.50 $6.00]
71. Show that the matrix equation has no solution.
represents the profit per unit. Find the product BA and
[11 1
1 ]
A=
1
0 [ 0
1 ] state what each entry of the matrix represents.

80. Manufacturing A corporation has three factories, Block Multiplication In Exercises 83 and 84, perform the
each of which manufactures acoustic guitars and block multiplication of matrices A and B. If matrices
electric guitars. In the matrix A and B are each partitioned into four submatrices

A= [
70 50 25
35 100 70 ] A= [
A11 A12
A21 A22
and B = ]
B11 B12
B21 B22 [ ]
aij represents the number of guitars of type i produced then you can block multiply A and B, provided the sizes of
at factory j in one day. Find the production levels when the submatrices are such that the matrix multiplications
production increases by 20%. and additions are defined.
81. Politics In the matrix
From
AB = [AA AA ] [BB
11
21
12
22
11
21
B12
B22 ]
=[ ]
A B +A B 11 11 12 21 A11B12 + A12B22
R D I A B +A B A21B12 + A22B22
21 11 22 21

[ ]
0.6 0.1 0.1 R

[ ]
P = 0.2 0.7 0.1 D 1 2 0

[ ]
To 1 2 0 0
0.2 0.2 0.8 I −1 1 0
83. A = 0 1 0 0 , B=
0 0 1
0 0 2 1
each entry pij (i ≠ j) represents the proportion of the 0 0 3
voting population that changes from party j to party i,

[ ] [ ]
and pii represents the proportion that remains loyal to 0 0 1 0 1 2 3 4
party i from one election to the next. Find and interpret 0 0 0 1 5 6 7 8
84. A = , B=
the product of P with itself. −1 0 0 0 1 2 3 4
82. Population The matrices show the numbers of 0 −1 0 0 5 6 7 8
people (in thousands) who lived in each region of
True or False? In Exercises 85 and 86, determine
the United States in 2010 and 2013. The regional
whether each statement is true or false. If a statement
populations are separated into three age categories.
is true, give a reason or cite an appropriate statement
(Source: U.S. Census Bureau)
from the text. If a statement is false, provide an example
2010 that shows the statement is not true in all cases or cite an
0–17 18–64 65+ appropriate statement from the text.

[ ]
Northeast 12,306 35,240 7830 85. (a) For the product of two matrices to be defined, the
Midwest 16,095 41,830 9051 number of columns of the first matrix must equal
the number of rows of the second matrix.
South 27,799 72,075 14,985
Mountain 5698 13,717 2710 (b) The system Ax = b is consistent if and only if b
Pacific 12,222 31,867 5901 can be expressed as a linear combination of the
columns of A, where the coefficients of the linear
2013 combination are a solution of the system.
0–17 18–64 65+ 86. (a) If A is an m × n matrix and B is an n × r matrix,

[ ]
then the product AB is an m × r matrix.
Northeast 12,026 35,471 8446
Midwest 15,772 41,985 9791 (b) The matrix equation Ax = b, where A is the
South 27,954 73,703 16,727 coefficient matrix and x and b are column matrices,
Mountain can be used to represent a system of linear equations.
5710 14,067 3104
Pacific 12,124 32,614 6636 87. The columns of matrix T show the coordinates of the
vertices of a triangle. Matrix A is a transformation matrix.
(a) The total population in 2010 was approximately
−1
309 million and the total population in 2013 was
about 316 million. Rewrite the matrices to give the
A= [010
, T=
1
1 ]
2
4
3
2 [ ]
information as percents of the total population. (a) Find AT and AAT. Then sketch the original
(b) Write a matrix that gives the changes in the percents triangle and the two transformed triangles. What
of the population in each region and age group from transformation does A represent?
2010 to 2013. (b)
A triangle is determined by AAT. Describe the
(c) Based on the result of part (b), which age group(s) transformation process that produces the triangle
show relative growth from 2010 to 2013? determined by AT and then the triangle determined
by T.

2.2 Properties of Matrix Operations

Use
the properties of matrix addition, scalar multiplication, and
zero matrices.
Use the properties of matrix multiplication and the identity matrix.
Find the transpose of a matrix.

Algebra of Matrices
In Section 2.1, you concentrated on the mechanics of the three basic matrix operations:
matrix addition, scalar multiplication, and matrix multiplication. This section begins to
develop the algebra of matrices. You will see that this algebra shares many (but not
all) of the properties of the algebra of real numbers. Theorem 2.1 lists several properties
of matrix addition and scalar multiplication.

Theorem 2.1 Properties of Matrix Addition

and Scalar Multiplication
If A, B, and C are m × n matrices, and c and d are scalars, then the properties
below are true.
1. A + B = B + A Commutative property of addition
2. A + (B + C) = (A + B) + C Associative property of addition
3. (cd)A = c(dA) Associative property of multiplication
4. 1A = A Multiplicative identity
5. c(A + B) = cA + cB Distributive property
6. (c + d)A = cA + dA Distributive property

proof
The proofs of these six properties follow directly from the definitions of matrix addition,
scalar multiplication, and the corresponding properties of real numbers. For example, to
prove the commutative property of matrix addition, let A = [aij ] and B = [bij ]. Then,
using the commutative property of addition of real numbers, write
A + B = [aij + bij ] = [bij + aij ] = B + A.
Similarly, to prove Property 5, use the distributive property (for real numbers) of
multiplication over addition to write
c(A + B) = [c(aij + bij )] = [caij + cbij ] = cA + cB.
The proofs of the remaining four properties are left as exercises. (See Exercises 61–64.)

The preceding section defined matrix addition as the sum of two matrices, making
it a binary operation. The associative property of matrix addition now allows you to
write expressions such as A + B + C as (A + B) + C or as A + (B + C). This same
reasoning applies to sums of four or more matrices.

Addition of More than Two Matrices

To obtain the sum of four matrices, add corresponding entries as shown below.

[12] + [−1
−1] [ 1] [ −3] [ −1]
+
0
+
2
=
2

One important property of the addition of real numbers is that the number 0
is the additive identity. That is, c + 0 = c for any real number c. For matrices,
a similar property holds. Specifically, if A is an m × n matrix and Omn is the
m × n matrix consisting entirely of zeros, then A + Omn = A. The matrix Omn is
a zero matrix, and it is the additive identity for the set of all m × n matrices. For
example, the matrix below is the additive identity for the set of all 2 × 3 matrices.

O23 = [00 0
0
0
0 ]
When the size of the matrix is understood, you may denote a zero matrix simply by O
or 0.
The properties of zero matrices listed below are relatively easy to prove, and their
proofs are left as an exercise. (See Exercise 65.)

reMArK
theOreM 2.2 Properties of Zero Matrices
Property 2 can be described
by saying that matrix −A is the If A is an m × n matrix and c is a scalar, then the properties below are true.
additive inverse of A. 1. A + Omn = A
2. A + (−A) = Omn
3. If cA = Omn, then c = 0 or A = Omn.

The algebra of real numbers and the algebra of matrices have many similarities.
For example, compare the two solutions below.
Real Numbers m × n Matrices
(Solve for x.) (Solve for X.)
x+a=b X+A=B
x + a + (−a) = b + (−a) X + A + (−A) = B + (−A)
x+0=b−a X+O=B−A
x=b−a X=B−A
Example 2 demonstrates the process of solving a matrix equation.

solving a Matrix equation

Solve for X in the equation 3X + A = B, where

−2
A= [10 3 ] and B= [−32 4
1 ]
.

sOlutiOn
Begin by solving the equation for X to obtain
3X = B − A
X = 13(B − A).
Now, using the matrices A and B, you have

X = 13 ([−32 41] − [10 −2

3 ])
= 13 [−42 −26]
[ ]
4
−3 2
= 2 .
3 − 23

Properties of Matrix Multiplication

The next theorem extends the algebra of matrices to include some useful properties of
matrix multiplication. The proof of Property 2 is below. The proofs of the remaining
REMARK properties are left as an exercise. (See Exercise 66.)
Note that no commutative
property of matrix
multiplication is listed in
Theorem 2.3 Properties of Matrix Multiplication
Theorem 2.3. The product If A, B, and C are matrices (with sizes such that the matrix products are defined),
AB may not be equal to the and c is a scalar, then the properties below are true.
product BA, as illustrated in 1. A(BC) = (AB)C Associative property of multiplication
Example 4 on the next page.
2. A(B + C) = AB + AC Distributive property
3. (A + B)C = AC + BC Distributive property
4. c(AB) = (cA)B = A(cB)

proof
To prove Property 2, show that the corresponding entries of matrices A(B + C) and
AB + AC are equal. Assume A has size m × n, B has size n × p, and C has size n × p.
Using the definition of matrix multiplication, the entry in the ith row and jth column
of A(B + C) is ai1(b1j + c1j ) + ai2(b2j + c2j) + . . . + ain(bnj + cnj ). Moreover, the
entry in the ith row and jth column of AB + AC is
(ai1b1j + ai2b2j + . . . + ainbnj ) + (ai1c1j + ai2c2j + . . . + aincnj ).
By distributing and regrouping, you can see that these two ijth entries are equal. So,
A(B + C) = AB + AC.

The associative property of matrix multiplication permits you to write such matrix
products as ABC without ambiguity, as demonstrated in Example 3.

Matrix Multiplication Is Associative

Find the matrix product ABC by grouping the factors first as (AB)C and then as A(BC).
Show that you obtain the same result from both processes.

[ ]
−1 0
−2
A=
1
2 [ −1
, B =
1
3] [ 0
−2
2
1 ]
, C = 3 1
2 4

solution
Grouping the factors as (AB)C, you have

[ ]
−1 0
−2
(AB)C = ([ 1
2 −1 ][ 1
3
0
−2
2
1 ]) 3 1
2 4

[ ]
−1 0
−5
= [ −1
4
2
0
3 ] 3 1 =
17
13 [ 4
14
.]
2 4
Grouping the factors as A(BC), you obtain the same result.

( [ ])
−1 0
−2
A(BC) = [ 1
2 ][
−1
1
3
0
−2
3
2
1 ] 1
2 4
−2
= [12 ][−73
−1
8
2
=] [
17
13 14
4
]

The next example shows that even when both products AB and BA are defined,
they may not be equal.

Noncommutativity of Matrix Multiplication

Show that AB and BA are not equal for the matrices

−1
A= [12 3
−1 ]
and B =
2
0 [ 2
. ]
solution
−1 −1
AB = [12 3
−1 ][20 2
=
2
4 ] [ 5
−4 ]
, BA =
2
0 [ 2 ][12 3
−1
=
0
4 ] [ 7
−2 ]
AB ≠ BA

Do not conclude from Example 4 that the matrix products AB and BA are never
equal. Sometimes they are equal. For example, find AB and BA for the matrices below.
−2
A= [11 2
1 ]
and B =
2 [ 4
−2 ]
You will see that the two products are equal. The point is that although AB and BA are
sometimes equal, AB and BA are usually not equal.
Another important quality of matrix algebra is that it does not have a general
cancellation property for matrix multiplication. That is, when AC = BC, it is not
necessarily true that A = B. Example 5 demonstrates this. (In the next section you will
see that, for some special types of matrices, cancellation is valid.)

An Example in Which Cancellation Is Not Valid

Show that AC = BC.

−2
A= [10 3
1 ]
, B =
2
2 [ 4
3 ]
, C =
1
−1 [ 2 ]
solution
−2 −2 −2 −2
AC = [10 3
1][−11 2
=
−1] [ 4
2 ]
, BC =
2
2 [ 4
3 ][−11 2
=
−1] [ ]
4
2
AC = BC, even though A ≠ B.

You will now look at a special type of square matrix that has 1’s on the main
diagonal and 0’s elsewhere.

[ ]
1 0 . . . 0
0 1 . . . 0
In =
⋮ ⋮ ⋮
0 0 . . . 1
n×n

For instance, for n = 1, 2, and 3,

[ ]
1 0 0
I1 = [1], I2 = [ 1
0
0
1 ]
, I3 = 0 1 0 .
0 0 1
When the order of the matrix is understood to be n, you may denote In simply as I.
As stated in Theorem 2.4 on the next page, the matrix In serves as the identity
for matrix multiplication; it is the identity matrix of order n. The proof of this theorem
is left as an exercise. (See Exercise 67.)

REMARK
Theorem 2.4 Properties of the Identity Matrix
Note that if A is a square matrix
of order n, then AIn = In A = A. If A is a matrix of size m × n, then the properties below are true.
1. AIn = A 2. Im A = A

Multiplication by an Identity Matrix

[ ][ [ ] [ ][ ] [ ]
3 −2 3 −2 1 0 0 −2 −2
a. 4 0
1
0
0
1
=] 4 0 b. 0 1 0 1 = 1
−1 1 −1 1 0 0 1 4 4

For repeated multiplication of square matrices, use the same exponential notation
used with real numbers. That is, A1 = A, A2 = AA, and for a positive integer k, Ak is
Ak = AA . . . A.
k factors

It is convenient also to define A0 = In (where A is a square matrix of order n). These

definitions allow you to establish the properties (1) A jAk = A j+k and (2) (A j)k = A jk,
where j and k are nonnegative integers.

Repeated Multiplication of a Square Matrix

−1
For the matrix A = [23 0
, ]
−1 −1 −1 1 −2 −1 −4 −1
A3 = ([23 0 ][23 0 ])[23 0
= ] [
6 −3 ][23 0
= ] [
3 −6
. ]
In Section 1.1, you saw that a system of linear equations has exactly one solution,
infinitely many solutions, or no solution. You can use matrix algebra to prove this.

Theorem 2.5 Number of Solutions of a Linear System

For a system of linear equations, precisely one of the statements below is true.
1. The system has exactly one solution.
2. The system has infinitely many solutions.
3. The system has no solution.

proof
Represent the system by the matrix equation Ax = b. If the system has exactly one
solution or no solution, then there is nothing to prove. So, assume that the system has
at least two distinct solutions x1 and x2. If you show that this assumption implies that
the system has infinitely many solutions, then the proof will be complete. When x1 and
x2 are solutions, you have Ax1 = Ax2 = b and A(x1 − x2 ) = O. This implies that the
(nonzero) column matrix xh = x1 − x2 is a solution of the homogeneous system of
linear equations Ax = O. So, for any scalar c,
A(x1 + cxh ) = Ax1 + A(cxh ) = b + c(Axh ) = b + cO = b.
Then x1 + cxh is a solution of Ax = b for any scalar c. There are infinitely many
possible values of c and each value produces a different solution, so the system has
infinitely many solutions.

the trAnsPOse Of A MAtrix

D I S COV E RY The transpose of a matrix is formed by writing its rows as columns. For example, if
1 2 A is the m × n matrix
Let A = [
4]
and
3

[ ]
a11 a12 a13 . . . a1n
3 5
B=[
1 −1]
. a21 a22 a23 . . . a2n
A= a31 a32 a33 . . . a3n
1. Find (AB) , A B , T T T ⋮ ⋮ ⋮ ⋮
and B A . T T am1 am2 am3 . . . amn
2. Make a conjecture Size: m × n
about the transpose then the transpose, denoted by AT, is the n × m matrix
of a product of two

[ ]
square matrices.
a11 a21 a31 . . . am1
a12 a22 a32 . . . am2
3. Select two other AT = a13 a23 a33 . . . am3 .
square matrices to
check your conjecture.
⋮ ⋮ ⋮ ⋮
a1n a2n a3n . . . amn
Size: n × m

transposes of Matrices

Find the transpose of each matrix.

reMArK

[ ] [ ] [ ]
1 2 3 1 2 0 0 1
Note that the square matrix in
part (c) is equal to its transpose.
a. A =
2
8 [] b. B = 4 5 6 c. C = 2 1 0 d. D = 2 4
Such a matrix is symmetric. 7 8 9 0 0 1 1 −1
A matrix A is symmetric when
sOlutiOn
A = AT. From this definition

[ ]
it should be clear that a 1 4 7
symmetric matrix must be a. AT = [2 8] b. BT = 2 5 8
square. Also, if A = [aij ] is
3 6 9
a symmetric matrix, then

[ ]
aij = aji for all i ≠ j. 1 2 0
c. C T = 2 1 0 d. DT = [01 2 1
4 −1 ]
0 0 1

theOreM 2.6 Properties of transposes

If A and B are matrices (with sizes such that the matrix operations are defined)
and c is a scalar, then the properties below are true.
1. (AT )T = A Transpose of a transpose
2. (A + B)T = AT + BT Transpose of a sum
3. (cA)T = c(AT ) Transpose of a scalar multiple
4. (AB)T = BTAT Transpose of a product

PrOOf
The transpose operation interchanges rows and columns, so Property 1 seems to make
sense. To prove Property 1, let A be an m × n matrix. Observe that AT has size n × m
and (AT )T has size m × n, the same as A. To show that (AT )T = A, you must show that
the ijth entries are the same. Let aij be the ijth entry of A. Then aij is the jith entry of
AT, and the ijth entry of (AT )T. This proves Property 1. The proofs of the remaining
properties are left as an exercise. (See Exercise 68.)

reMArK Properties 2 and 4 can be generalized to cover sums or products of any finite
Remember that you reverse the number of matrices. For instance, the transpose of the sum of three matrices is
order of multiplication when (A + B + C)T = AT + BT + C T and the transpose of the product of three matrices is
forming the transpose of a (ABC)T = C TBTAT.
product. That is, the transpose
of AB is (AB )T = BTAT and is
usually not equal to ATBT.
finding the transpose of a Product

See LarsonLinearAlgebra.com for an interactive version of this type of example.

Show that (AB)T and BTAT are equal.

[ ] [ ]
2 1 −2 3 1
A = −1 0 3 and B = 2 −1
0 −2 1 3 0

sOlutiOn

[ ][ ] [ ]
2 1 −2 3 1 2 1
AB = −1 0 3 2 −1 = 6 −1
0 −2 1 3 0 −1 2
6 −1
(AB)T = [21 −1 2 ]

[ ]
2 −1 0
6 −1
BTAT =
3 2
1 −1[ 3
0 ] 1 0 −2 =
2
1 −1 2 [ ]
−2 3 1
(AB)T = BTAT

the Product of a Matrix and its transpose

[ ]
1 3
reMArK For the matrix A = 0 −2 , find the product AAT and show that it is symmetric.
−2 −1
The property demonstrated in
Example 10 is true in general.
sOlutiOn
That is, for any matrix A, the

[ ][ [ ]
matrix AAT is symmetric. The 1 3 10 −6 −5
0 −2
matrix ATA is also symmetric.
You are asked to prove these
AAT = 0 −2
1
3 −2 −1
= −6 4 2 ]
−2 −1 −5 2 5
properties in Exercise 69.
It follows that AAT = (AAT )T, so AAT is symmetric.

Information retrieval systems such as Internet search engines

lineAr make use of matrix theory and linear algebra to keep track of
AlgebrA information. To illustrate, consider a simplified example. You
APPlieD could represent the occurrences of m available keywords in
a database of n documents with A, an m × n matrix in which
an entry is 1 when the keyword occurs in the document
and 0 when it does not occur in the document. You could
represent a search with the m × 1 column matrix x, in which
a 1 entry represents a keyword you are searching and 0
represents a keyword you are not searching. Then, the n × 1
matrix product ATx would represent the number of keywords
in your search that occur in each of the n documents. For
a discussion on the PageRank algorithm that is used in
Google’s search engine, see Section 2.5 (page 86).
Gunnar Pippel/Shutterstock.com

2.2 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

Evaluating an Expression In Exercises 1–6, evaluate Associativity of Matrix Multiplication In Exercises 23

the expression. and 24, find the matrix product ABC by (a) grouping
−5 −10 −8 the factors as (AB)C, and (b) grouping the factors as
1. [ 3 −6
0
+
7
−2 ] [ 1
−1] [
+
14 6] A(BC). Show that you obtain the same result from
both processes.
−11 −7
2. [
0] [ −3 −1] [ 2 −1]
6 8 0 5
+ +
−1 23. A =
1
3
2
4 [
, B=
0
2
1
3 ]
, C=
3
0
0
1 [ ] [ ]
−4 −2
3. 4 ([ 0
0
2
1
3 ] [
−
2
3
1
−6 0 ]) 24. A =
−4 2
[, B=
1 −5 0
, ] [ ]
1 −3 −2 3 3
1
4. 2 ([5 −2 4 0] + [14 6 −18 9])

[ ]
−3 4
−3 −6 −4
5. −3 ([ 0
7 2
+
8] [ 3
1 ]) −2
4
7 [ −9 ] C= 0
−1
1
1

[ ] ([ ] [ ])
4 11 −5 −1 7 5
1 Noncommutativity of Matrix Multiplication In
6. − −2 −1 + 3 4 + −9 −1
6 Exercises 25 and 26, show that AB and BA are not equal
9 3 0 13 6 −1 for the given matrices.
−2
Operations with Matrices In Exercises 7–12, perform
the operations, given a = 3, b = −4, and
25. A =
0
1
3[, B=
−1
4 0
2 ] [ ]
1
[2 0
] 1 0
[0
] [ ]
1 1 1 1
A=
3 4
, B=
−1 2
, O=
0 0
. 26. A = [ 4
1
2
2
1
2
]
, B= [
2
1
2
2
1
4
]
7. aA + bB 8. A + B
9. ab(B) 10. (a + b)B Equal Matrix Products In Exercises 27 and 28, show
that AC = BC, even though A ≠ B.
11. (a − b)(A − B) 12. (ab)O

13. Solve for X in the equation, given

27. A =
0
0
1
1 [
, B=
1
1
0
0
, C=
2
2] [ ] [ ]
3
3

[ ] [ ]
−6

[ ] [ ]
−4 0 1 2 1 2 3 4 3
A= 1 −5 and B = −2 1 . 28. A = 0 5 4 , B= 5 4 4 ,
−3 2 4 4 3 −2 1 −1 0 1

[ ]
(a) 3X + 2A = B (b) 2A − 5B = 3X 0 0 0
(c) X − 3A + 2B = O (d) 6X − 4A − 3B = O C= 0 0 0
4 −2 3
14. Solve for X in the equation, given
Zero Matrix Product In Exercises 29 and 30, show that

[ ] [ ]
−2 −1 0 3
A= 1 0 and B= 2 0 . AB = O, even though A ≠ O and B ≠ O.
1 −1
3 −4 −4 −1 29. A =
3
4
3
4 [
and B =
−1 ]1 [ ]
(a) X = 3A − 2B (b) 2X = 2A − B
1 −2
(c) 2X + 3A = B (d) 2A + 4B = −2X 30. A =
2
2
4
4 [
and B =
−12 ]1 [ ]
Operations with Matrices In Exercises 15–22, Operations with Matrices In Exercises 31–36,
perform the operations, given c = −2 and perform the operations when
A=
1
0 [
2
1 −1
3
,B=
−1
1 3
2
,C= ] 0
−1
1
0
. [ ] [ ] A= [10 −12].
15. c(BA) 16. c(CB) 31. IA 32. AI
17. B(CA) 18. C(BC) 33. A(I + A) 34. A + IA
19. (B + C)A 20. B(C + O) 35. A2 36. A4
21. cB(C + C) 22. B(cA)

Writing In Exercises 37 and 38, explain why the Finding an nth Root of a Matrix In Exercises 53
formula is not valid for matrices. Illustrate your and 54, find the nth root of the matrix B. An nth root of
argument with examples. a matrix B is a matrix A such that An = B.
37. (A + B)(A − B) = A2 − B2
38. (A + B)(A + B) = A2 + 2AB + B2
53. B =
9
0
0
4 [
, n=2 ]

[ ]
8 0 0
Finding the Transpose of a Matrix In Exercises 39
54. B = 0 −1 0 , n=3
and 40, find the transpose of the matrix.
0 0 27

[ ] [ ]
1 −2 6 −7 19
39. D = −3 4 40. D = −7 0 23 True or False? In Exercises 55 and 56, determine
5 −1 19 23 −32 whether each statement is true or false. If a statement
is true, give a reason or cite an appropriate statement
Finding the Transpose of a Product of Two Matrices from the text. If a statement is false, provide an example
In Exercises 41–44, verify that (AB)T = BTAT. that shows the statement is not true in all cases or cite an

[ ]
−3 0 appropriate statement from the text.
−1 1 −2
41. A = [
2 0 1
and B = 1 ]2 55. (a) Matrix addition is commutative.
1 −1
(b) The transpose of the product of two matrices equals
42. A = [10 2
−2 ] and B= [−32 −1
1 ] the product of their transposes; that is, (AB)T = ATBT.
(c) For any matrix C the matrix CC T is symmetric.

[ ]
2 1 56. (a) Matrix multiplication is commutative.
43. A = 0 1 and B= [20 3
4
1
−1 ] (b) If the matrices A, B, and C satisfy AB = AC, then
−2 1 B = C.
(c) The transpose of the sum of two matrices equals the

[ ] [ ]
2 1 −1 1 0 −1
44. A = 0 1 3 and B= 2 1 −2 sum of their transposes.
4 0 2 0 1 3 57. Consider the matrices below.

[] [] [ ] []
Multiplication with the Transpose of a Matrix 1 1 2 1
In Exercises 45–48, find (a) ATA and (b) AAT. Show that X= 0 , Y= 1 , Z = −1 , W = 1
each of these products is symmetric. 1 0 3 1

[ ]
1 −1
(a) Find scalars a and b such that Z = aX + bY.
45. A =
4
0 [ 2
2 −1
1
46. A = 3 ] 4
(b) Show that there do not exist scalars a and b such
0 −2
that W = aX + bY.

[ ]
0 −4 3 2
(c) Show that if aX + bY + cW = O, then a = 0, b = 0,
8 4 0 1
47. A = and c = 0.
−2 3 5 1
(d) Find scalars a, b, and c, not all equal to zero, such
0 0 −3 2
that aX + bY + cZ = O.

[ ]
4 −3 2 0
2 0 11 −1 58. CAPSTONE In the matrix equation
48. A = −1 −2 0 3
aX + A(bB) = b(AB + IB)
14 −2 12 −9
X, A, B, and I are square matrices, and a and b are
6 8 −5 4
nonzero scalars. Justify each step in the solution
Finding a Power of a Matrix In Exercises 49–52, find below.
the power of A for the matrix aX + (Ab)B = b(AB + B)

[ ]
1 0 0 0 0 aX + bAB = bAB + bB
0 −1 0 0 0 aX + bAB + (−bAB) = bAB + bB + (−bAB)
A= 0 0 1 0 0 . aX = bAB + bB + (−bAB)
0 0 0 −1 0 aX = bAB + (−bAB) + bB
0 0 0 0 1 aX = bB
49. A16 50. A17 b
X= B
51. A19 52. A20 a

Polynomial Function In Exercises 59 and 60, find f (A) 70. Proof Let A and B be two n × n symmetric matrices.
using the definition below. (a) Give an example to show that the product AB is not
If f (x) = a0 + a1x + a2 x2 + . . . + an xn is a polynomial necessarily symmetric.
function, then for a square matrix A, (b) Prove that the product AB is symmetric if and only
f (A) = a0 I + a1A + a2A2 + . . . + anAn. if AB = BA.

59. f (x) = 2 − 5x + x2, A= [24 0

5 ] Symmetric and Skew-Symmetric Matrices In
Exercises 71–74, determine whether the matrix is
symmetric, skew-symmetric, or neither. A square matrix

[ ]
2 1 −1
60. f (x) = −10 + 5x − 2x2 + x3, A = 1 0 2 is skew-symmetric when AT = −A.
−1 1 3 71. A =
−2
0 2
0 [ 72. A =
2
1 ]1
3 [ ]
61. Guided Proof Prove the associative property of

[ ] [ ]
0 2 1 0 2 −1
matrix addition: A + (B + C) = (A + B) + C.
73. A = 2 0 3 74. A = −2 0 −3
Getting Started: To prove that A + (B + C) and 1 3 0 1 3 0
(A + B) + C are equal, show that their corresponding
entries are equal. 75. Proof Prove that the main diagonal of a skew-
symmetric matrix consists entirely of zeros.
(i) Begin your proof by letting A, B, and C be m × n
matrices. 76. Proof Prove that if A and B are n × n skew-symmetric
matrices, then A + B is skew-symmetric.
(ii) Observe that the ijth entry of B + C is bij + cij.
77. Proof Let A be a square matrix of order n.
(iii) Furthermore, the ijth entry of A + (B + C) is
(a) Show that 12(A + AT ) is symmetric.
aij + (bij + cij).
(b) Show that 12(A − AT ) is skew-symmetric.
(iv) Determine the ijth entry of (A + B) + C.
(c) Prove that A can be written as the sum of a
62. Proof Prove the associative property of multiplication:
symmetric matrix B and a skew-symmetric matrix C,
(cd)A + c(dA).
A = B + C.
63. Proof Prove that the scalar 1 is the identity for scalar
(d) Write the matrix below as the sum of a symmetric
multiplication: 1A = A.
matrix and a skew-symmetric matrix.
64. Proof Prove the distributive property:

[ ]
2 5 3
(c + d)A = cA + dA.
A = −3 6 0
65. Proof Prove Theorem 2.2. 4 1 1
66. Proof Complete the proof of Theorem 2.3.
78. Proof Prove that if A is an n × n matrix, then A − AT
(a) Prove the associative property of multiplication: is skew-symmetric.
A(BC) = (AB)C. 79. Consider matrices of the form
(b) Prove the distributive property:

[ ]
(A + B)C = AC + BC. 0 a12 a13 . . . a1n
0 0 a23 . . . a2n
(c) Prove the property:
A= ⋮ ⋮ ⋮ ⋮ .
c(AB) = (cA)B = A(cB). 0 0 0 . . . an−1, n
67. Proof Prove Theorem 2.4. 0 0 0 . . . 0
68. Proof Prove Properties 2, 3, and 4 of Theorem 2.6.
(a) Write a 2 × 2 matrix and a 3 × 3 matrix in the form
69. Guided Proof Prove that if A is an m × n matrix,
of A.
then AAT and ATA are symmetric matrices.
(b) Use a graphing utility to raise each of the matrices
Getting Started: To prove that AAT is symmetric, you need
to higher powers. Describe the result.
to show that it is equal to its transpose, (AAT )T = AAT.
(c) Use the result of part (b) to make a conjecture
(i)
Begin your proof with the left-hand matrix
about powers of A when A is a 4 × 4 matrix. Use a
expression (AAT )T.
graphing utility to test your conjecture.
(ii)
Use the properties of the transpose operation
(d) Use the results of parts (b) and (c) to make a
to show that (AAT )T can be simplified to equal the
conjecture about powers of A when A is an
right-hand expression, AAT.
n × n matrix.
(iii) Repeat this analysis for the product ATA.

2.3 The Inverse of a Matrix

Find the inverse of a matrix (if it exists).
Use properties of inverse matrices.
Use an inverse matrix to solve a system of linear equations.

Matrices and Their Inverses

Section 2.2 discussed some of the similarities between the algebra of real numbers and
the algebra of matrices. This section further develops the algebra of matrices to include
the solutions of matrix equations involving matrix multiplication. To begin, consider
the real number equation ax = b. To solve this equation for x, multiply both sides of
the equation by a−1 (provided a ≠ 0).
ax = b
(a−1a)x = a−1b
(1)x = a−1b
x = a−1b
The number a−1 is the multiplicative inverse of a because a−1a = 1 (the identity
element for multiplication). The definition of the multiplicative inverse of a matrix is
similar.

Definition of the Inverse of a Matrix

An n × n matrix A is invertible (or nonsingular) when there exists an n × n
matrix B such that
AB = BA = In
where In is the identity matrix of order n. The matrix B is the (multiplicative)
inverse of A. A matrix that does not have an inverse is noninvertible (or
singular).

Nonsquare matrices do not have inverses. To see this, note that if A is of size
m × n and B is of size n × m (where m ≠ n), then the products AB and BA are of
different sizes and cannot be equal to each other. Not all square matrices have inverses.
(See Example 4.) The next theorem, however, states that if a matrix does have an
inverse, then that inverse is unique.

Theorem 2.7 Uniqueness of an Inverse Matrix

If A is an invertible matrix, then its inverse is unique. The inverse of A is
denoted by A−1.

proof
If A is invertible, then it has at least one inverse B such that
AB = I = BA.
Assume that A has another inverse C such that
AC = I = CA.
Demonstrate that B and C are equal, as shown on the next page.

AB = I
C(AB) = CI
(CA)B = C
IB = C
B=C
Consequently B = C, and it follows that the inverse of a matrix is unique.

The inverse A−1 of an invertible matrix A is unique, so you can call it the inverse
of A and write AA−1 = A−1A = I.

The Inverse of a Matrix

Show that B is the inverse of A, where

REMARK A= [−1
−1
2
1 ] and B= [11 −2
−1
. ]
Recall that it is not always true
that AB = BA, even when both solution
products are defined. If A and
Using the definition of an inverse matrix, show that B is the inverse of A by showing
B are both square matrices and
that AB = I = BA.
AB = In , however, then it can
be shown that BA = In . (The
proof of this is omitted.) So, AB = [−1
−1
2
1 ][11 −2
−1
=
−1 + 2
] [
−1 + 1
2−2
2−1
=
1
0 ] [ 0
1 ]
in Example 1, you need only
−2
check that AB = I2.
BA = [11 −1 ][−1
−1
2
1
=
−1 + 2
] [
−1 + 1
2−2
2−1
=
1
0 ] [ 0
1 ]
The next example shows how to use a system of equations to find the inverse of
a matrix.

Finding the Inverse of a Matrix

Find the inverse of the matrix

A= [−11 4
−3
. ]
solution
To find the inverse of A, solve the matrix equation AX = I for X.

[−11 4
−3 ][xx 11
21
x12
x22
= ] [
1
0
0
1 ]
+ 4x21 x12 + 4x22
[−xx 11
11 − 3x21 −x12 − 3x22
=
1
0 ] [ 0
1 ]
Equating corresponding entries, you obtain two systems of linear equations.
x11 + 4x21 = 1 x12 + 4x22 = 0
−x11 − 3x21 = 0 −x12 − 3x22 = 1
Solving the first system, you find that x11 = −3 and x21 = 1. Similarly, solving the
second system, you find that x12 = −4 and x22 = 1. So, the inverse of A is

X = A−1 = [−31 −4
1
. ]
Use matrix multiplication to check this result.

Generalizing the method used to solve Example 2 provides a convenient method

for finding an inverse. Note that the two systems of linear equations
x11 + 4x21 = 1 x12 + 4x22 = 0
−x11 − 3x21 = 0 −x12 − 3x22 = 1
have the same coefficient matrix. Rather than solve the two systems represented by

[−11 4
−3
1
0 ] and [−11 4
−3
0
1 ]
separately, solve them simultaneously by adjoining the identity matrix to the coeficient
matrix to obtain

[−11 4
−3
1
0
0
1 ]
.

By applying Gauss-Jordan elimination to this matrix, solve both systems with a single
elimination process, as shown below.

[10 4
1
1
1
0
1 ] R2 + R1 → R2
−3 −4 R1 + (−4)R2 → R1
[10 0
1 1 1 ]
Applying Gauss-Jordan elimination to the “doubly augmented” matrix [A I ], you
obtain the matrix [I A−1].
−3 −4
[−11 4
−3
1
0
0
1 ] [10 0
1 1 1 ]
A I I A−1
This procedure (or algorithm) works for an arbitrary n × n matrix. If A cannot be row
reduced to In, then A is noninvertible (or singular). This procedure will be formally
justified in the next section, after introducing the concept of an elementary matrix. For
now, a summary of the algorithm is shown below.

finding the inverse of a Matrix by Gauss-Jordan elimination

Let A be a square matrix of order n.
1. Write the n × 2n matrix that consists of A on the left and the n × n identity
matrix I on the right to obtain [A I ]. This process is called adjoining
matrix I to matrix A.
2. If possible, row reduce A to I using elementary row operations on the entire
matrix [A I ]. The result will be the matrix [I A−1]. If this is not
possible, then A is noninvertible (or singular).
3. Check your work by multiplying to see that AA−1 = I = A−1A.

Recall Hooke’s law, which states that for relatively small

linear deformations of an elastic object, the amount of deflection is
alGeBra directly proportional to the force causing the deformation. In
applied a simply supported elastic beam subjected to multiple forces,
deflection d is related to force w by the matrix equation
d = Fw
where F is a flexibility matrix whose entries depend on the
material of the beam. The inverse of the flexibility matrix,
F −1, is the stiffness matrix. In Exercises 61 and 62, you are
asked to find the stiffness matrix F −1 and the force matrix w
for a given set of flexibility and deflection matrices.
nostal6ie/Shutterstock.com

finding the inverse of a Matrix

See LarsonLinearAlgebra.com for an interactive version of this type of example.

Find the inverse of the matrix.

[ ]
1 −1 0
A= 1 0 −1
−6 2 3

solUtion
Begin by adjoining the identity matrix to A to form the matrix

[ ]
1 −1 0 1 0 0
[A I] = 1 0 −1 0 1 0 .
−6 2 3 0 0 1
Use elementary row operations to obtain the form
[I A−1]
as shown below.

[ ]
1 −1 0 1 0 0
0 1 −1 −1 1 0 R2 + (−1)R1 → R2
−6 2 3 0 0 1

[ ]
1 −1 0 1 0 0
0 1 −1 −1 1 0
0 −4 3 6 0 1 R3 + (6)R1 → R3

[ ]
1 −1 0 1 0 0
0 1 −1 −1 1 0
technoloGy 0 0 −1 2 4 1 R3 + (4)R2 → R3

[ ]
Many graphing utilities and 1 −1 0 1 0 0
software programs can 0 1 −1 −1 1 0
find the inverse of a square 0 0 1 −2 −4 −1 (−1)R3 → R3
matrix. When you use a

[ ]
graphing utility, you may 1 −1 0 1 0 0
see something similar to the 0 1 0 −3 −3 −1 R2 + R3 → R2
screen below for Example 3.
0 0 1 −2 −4 −1
The technology Guide at

[ ]
CengageBrain.com can help 1 0 0 −2 −3 −1 R1 + R2 → R1
you use technology to find 0 1 0 −3 −3 −1
the inverse of a matrix.
0 0 1 −2 −4 −1
A The matrix A is invertible, and its inverse is
[[1 -1 0 ]
[1 0 -1]

[ ]
[-6 2 3 ]] −2 −3 −1
A-1 A−1 = −3 −3 −1 .
[[-2 -3 -1]
[-3 -3 -1] −2 −4 −1
[-2 -4 -1]]
Confirm this by showing that
AA−1 = I = A−1A.

The process shown in Example 3 applies to any n × n matrix A and will find the
inverse of A, if it exists. When A has no inverse, the process will also tell you that. The
next example applies the process to a singular matrix (one that has no inverse).

A Singular Matrix

Show that the matrix has no inverse.

[ ]
1 2 0
A= 3 −1 2
−2 3 −2

solution
Adjoin the identity matrix to A to form

[ ]
1 2 0 1 0 0
[A I] = 3 −1 2 0 1 0
−2 3 −2 0 0 1
and apply Gauss-Jordan elimination to obtain

[ ]
1 2 0 1 0 0
0 −7 2 −3 1 0 .
0 0 0 −1 1 1
Note that the “A portion” of the matrix has a row of zeros. So it is not possible to
rewrite the matrix [A I ] in the form [I A−1]. This means that A has no inverse, or is
noninvertible (or singular).

Using Gauss-Jordan elimination to find the inverse of a matrix works well (even
as a computer technique) for matrices of size 3 × 3 or greater. For 2 × 2 matrices,
however, you can use a formula for the inverse rather than Gauss-Jordan elimination.
If A is a 2 × 2 matrix

REMARK A= [ac b
d ]
The denominator ad − bc is
called the determinant of A. then A is invertible if and only if ad − bc ≠ 0. Moreover, if ad − bc ≠ 0, then the
You will study determinants inverse is
in detail in Chapter 3.
−b
A−1 =
1 d
ad − bc −c [ a
. ]
Finding Inverses of 2 × 2 Matrices

If possible, find the inverse of each matrix.

−1 −1
a. A = [−23 2 ] b. B = [−63 2 ]
solution
a. For the matrix A, apply the formula for the inverse of a 2 × 2 matrix to obtain
ad − bc = (3)(2) − (−1)(−2) = 4. This quantity is not zero, so A is invertible.
Form the inverse by interchanging the entries on the main diagonal, changing the
signs of the other two entries, and multiplying by the scalar 14, as shown below.

[ ]
1 1
A−1 = 1
4 [ 2
2
1
3]=
2
1
2
4
3
4

or the matrix B, you have ad − bc = (3)(2) − (−1)(−6) = 0, which means that

b. F
B is noninvertible.

properties of inverses
Theorem 2.8 below lists important properties of inverse matrices.

theoreM 2.8 properties of inverse Matrices

If A is an invertible matrix, k is a positive integer, and c is a nonzero scalar, then
A−1, Ak, cA, and AT are invertible and the statements below are true.
1. (A−1)−1 = A 2. (Ak)−1 = A−1A−1 . . . A−1 = (A−1)k

k factors
1
3. (cA)−1 = A−1 4. (AT )−1 = (A−1)T
c

proof
The key to the proofs of Properties 1, 3, and 4 is the fact that the inverse of a matrix is
unique (Theorem 2.7). That is, if BC = CB = I, then C is the inverse of B.
Property 1 states that the inverse of A−1 is A itself. To prove this, observe that
A−1A = AA−1 = I, which means that A is the inverse of A−1. Thus, A = (A−1)−1.
1
Similarly, Property 3 states that A−1 is the inverse of (cA), c ≠ 0. To prove this,
c
use the properties of scalar multiplication given in Theorems 2.1 and 2.3.

(cA) (1cA ) = (c1c )AA

−1 −1 = (1)I = I

(1cA )(cA) = (1cc)A

−1 −1A = (1)I = I

1 1
So A−1 is the inverse of (cA), which implies that (cA)−1 = A−1. Properties 2 and 4
c c
are left for you to prove. (See Exercises 63 and 64.)

For nonsingular matrices, the exponential notation used for repeated multiplication
of square matrices can be extended to include exponents that are negative integers. This
may be done by defining A−k to be

A−k = A−1A−1 . . . A−1 = (A−1)k.

k factors

Using this convention you can show that the properties AjAk = Aj+k and (Aj )k = Ajk are
true for any integers j and k.

DISCO VERY
−1
Let A = [11 2
3 ] and B = [21 −1
. ]
1. Find (AB )−1, A−1B −1, and B −1A−1.

2. Make a conjecture about the inverse of a product of two nonsingular

matrices. Then select two other nonsingular matrices of the same
order and see whether your conjecture holds.

See LarsonLinearAlgebra.com for an interactive version of this type of exercise.

The Inverse of the Square of a Matrix

Compute A−2 two different ways and show that the results are equal.

A= [12 1
4 ]
solution
One way to find A−2 is to find (A2)−1 by squaring the matrix A to obtain

A2 = [103 5
18 ]
and using the formula for the inverse of a 2 × 2 matrix to obtain

[ ]
9
−5 − 54
(A2)−1 = 14 [ 18
−10 3
= ]
2

− 52
3
4
.

Another way to find A−2 is to find (A−1)2 by finding A−1

[ ]
−1 2 − 12
A−1 = 12 [ 4
−2 1
=
−1
] 1
2

and then squaring this matrix to obtain

[ ]
9
2 − 54
(A−1)2 = .
− 52
3
4

Note that both methods produce the same result.

The next theorem gives a formula for computing the inverse of a product of
two matrices.

Theorem 2.9 The Inverse of a Product

If A and B are invertible matrices of order n, then AB is invertible and
(AB)−1 = B−1A−1.

proof
To show that B−1A−1 is the inverse of AB, you need only show that it conforms to the
definition of an inverse matrix. That is,
(AB)(B−1A−1) = A(BB−1)A−1 = A(I)A−1 = (AI)A−1 = AA−1 = I.
In a similar way, (B−1A−1)(AB) = I. So, AB is invertible and its inverse is B−1A−1.

Theorem 2.9 states that the inverse of a product of two invertible matrices is the
product of their inverses taken in the reverse order. This can be generalized to include
the product of more than two invertible matrices:
(A1A2A3 . . . An)−1 = An−1 . . . A3−1A2−1A1−1.
(See Example 4 in Appendix.)

Finding the Inverse of a Matrix Product

Find (AB)−1 for the matrices

[ ] [ ]
1 3 3 1 2 3
A= 1 4 3 and B= 1 3 3
1 3 4 2 4 3

using the fact that A−1 and B−1 are

[ ] [ ]
7 −3 −3 1 −2 1
REMARK A−1 = −1 1 0 and B−1 = −1 1 0 .
2
Note that you reverse the order −1 0 1 3 0 − 13
of multiplication to find the
inverse of AB. That is,
solution
(AB )−1 = B −1A−1, and the
inverse of AB is usually not Using Theorem 2.9 produces
equal to A−1B −1.
(AB)−1 = B−1A−1

[ ][ ]
1 −2 1 7 −3 −3
= −1 1 0 −1 1 0
2
3 0 − 13 −1 0 1

[ ]
8 −5 −2
= −8 4 3 .
5 −2 − 73

One important property in the algebra of real numbers is the cancellation property.
That is, if ac = bc (c ≠ 0), then a = b. Invertible matrices have similar cancellation
properties.

Theorem 2.10 Cancellation Properties

If C is an invertible matrix, then the properties below are true.
1. If AC = BC, then A = B. Right cancellation property
2. If CA = CB, then A = B. Left cancellation property

proof
To prove Property 1, use the fact that C is invertible and write
AC = BC
(AC)C−1 = (BC)C−1
A(CC−1) = B(CC−1)
AI = BI
A = B.
The second property can be proved in a similar way. (See Exercise 65.)

Be sure to remember that Theorem 2.10 can be applied only when C is an

invertible matrix. If C is not invertible, then cancellation is not usually valid. For
instance, Example 5 in Section 2.2 gives an example of a matrix equation AC = BC
in which A ≠ B, because C is not invertible in the example.

Systems of Equations
For square systems of equations (those having the same number of equations as variables),
you can use the theorem below to determine whether the system has a unique solution.

Theorem 2.11 Systems of Equations with Unique Solutions

If A is an invertible matrix, then the system of linear equations Ax = b has a
unique solution x = A−1b.

proof
The matrix A is nonsingular, so the steps shown below are valid.
Ax = b
A−1Ax = A−1b
Ix = A−1b
x = A−1b
This solution is unique because if x1 and x2 were two solutions, then you could apply
the cancellation property to the equation Ax1 = b = Ax2 to conclude that x1 = x2.

One use of Theorem 2.11 is in solving several systems that all have the same
coefficient matrix A. You could find the inverse matrix once and then solve each
system by computing the product A−1b.

Solving Systems of Equations

Using an Inverse Matrix
Use an inverse matrix to solve each system.
a. 2x + 3y + z = −1 b. 2x + 3y + z = 4 c. 2x + 3y + z = 0
3x + 3y + z = 1 3x + 3y + z = 8 3x + 3y + z = 0
2x + 4y + z = −2 2x + 4y + z = 5 2x + 4y + z = 0
solution

[ ]
2 3 1
First note that the coefficient matrix for each system is A = 3 3 1 .
2 4 1

[ ]
−1 1 0
Using Gauss-Jordan elimination, A−1 = −1 0 1 .
6 −2 −3

[ ][ ] [ ]
−1 1 0 −1 2
The solution is x = 2,
a. x = A−1b = −1 0 1 1 = −1
y = −1, and z = −2.
6 −2 −3 −2 −2

[ ][ ] [ ]
−1 1 0 4 4
b. x = A−1b = −1 0 1 8 = 1 The solution is x = 4,
y = 1, and z = −7.
6 −2 −3 5 −7

[ ][ ] [ ]
−1 1 0 0 0
The solution is trivial:
c. x = A−1b = −1 0 1 0 = 0
x = 0, y = 0, and z = 0.
6 −2 −3 0 0

2.3 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

[ ] [ ]
The Inverse of a Matrix In Exercises 1–6, show that B −8 0 0 0 1 0 0 0
is the inverse of A. 0 1 0 0 0 2 0 0
−1 25. 26.
1. A = [ 2
5 ] B = [−53
1
3
,
2 ] 0
0
0
0
0
0
0
−5
0
0
0
0
−2
0
0
3
1 −1
2. A = [
2]
, B=[
1]
2 1

[ ] [ ]
1 −2 −1 −2 4 8 −7 14
−1 1
3 −5 −2 −3 2 5 −4 6
27. 28.
[
−2
]
1
3. A = [13 2
4 ]
, B= 3
2 − 12
2
−1
−5
4
−2
4
−5
11
0
3
2
6
1
−5
−7
10

[ ]
[ ] [ ]
3 1
−1 1 0 3 0 1 3 −2 0
4. A = [ 1
2 3
, ] B=
5

− 25
5
1
5 29.
0 2 0 4
30.
0 2 4 6
1 0 3 0 0 0 −2 1

[ ] [ ]
−2 2 3 −4 −5 3 0 2 0 4 0 0 0 5
5. A = 1 −1 0 , B = 13 −4 −8 3
0 1 4 1 2 0 Finding the Inverse of a 2 × 2 Matrix In Exercises
31–36, use the formula on page 66 to find the inverse of

[ ] [ ]
2 −17 11 1 1 2
the 2 × 2 matrix (if it exists).
6. A = −1 11 −7 , B= 2 4 −3
1 −2
0 3 −2 3 6 −5 31. [ 2
−1
3
5 ] 32.
−3 2 [ ]
Finding the Inverse of a Matrix In Exercises 7–30, −4 −6 −12
33. [
3]
34. [ ]
3
find the inverse of the matrix (if it exists). 2 5 −2
2 −2
[ 2 0
] [ ]
[ ] [ ]
7
− 34 1 9
7. 8. 2 −4 4
0 3 2 2 35. 36.
1 4 5 8
1 −2
9. [
7]
10. [
2 −3]
1 2 5 5 3 9

3 Finding the Inverse of the Square of a Matrix

−7 33 −1 In Exercises 37–40, compute A−2 two different ways and
11. [
4 −19]
12. [
3 −3]
1
show that the results are equal.
0 −2
[ 2
] 7
[ ]
[ ] [ ]
1 1 1 1 2 2 37. A = 38. A =
−1 3 −5 6
13. 3 5 4 14. 3 7 9

[ ] [ ]
3 6 5 −1 −4 −7 −2 0 0 6 0 4
39. A = 0 1 0 40. A = −2 7 −1

[ ] [ ]
1 2 −1 10 5 −7
0 0 3 3 1 2
15. 3 7 −10 16. −5 1 4
7 16 −21 3 2 −2 Finding the Inverses of Products and Transposes
In Exercises 41–44, use the inverse matrices to find

[ ] [ ]
1 1 2 3 2 5 (a) (AB)−1, (b) (AT )−1, and (c) (2A)−1.
17. 3 1 0 18. 2 2 4 7 −3
−2 0 3 −4 4 0 41. A−1 =
−7
2 5
6 [
, B−1 =
2 0 ] [ ]

[ ]
− 56 1 11

[ ] [ ]
2 1 5 2
−7

[ ]
2 0 0 3 6 7 11 11
2 42. A−1 = 3 2 , B−1 = 3 1
19. 0 3 0 20. 0 3 2 7 7 11 − 11
0 0 5 − 12 − 52

[ ] [ ]
1 1 − 12 3
2 4
5
4 2

[ ] [ ]
0.6 0 −0.3 0.1 0.2 0.3 43. A−1 = 3
2
1
2 −2 , B−1 = − 34 2
1
4
21. 0.7 −1 0.2 22. −0.3 0.2 0.2 1 1 1 1
4 1 2 4 2 2
1 0 −0.9 0.5 0.5 0.5

[ ] [ ]
1 −4 2 6 5 −3

[ ] [ ]
1 0 0 1 0 0
44. A−1 = 0 1 3 , B−1 = −2 4 −1
23. 3 4 0 24. 3 0 0
4 2 1 1 3 4
2 5 5 2 5 5

solving a system of equations Using an inverse singular Matrix In Exercises 55 and 56, find x such
In Exercises 45–48, use an inverse matrix to solve each that the matrix is singular.
system of linear equations.
45. (a) x + 2y = −1 46. (a) 2x − y = −3
55. A =
4
−2 −3
x
[ 56. A =
x
−3 ]
2
4 [ ]
x − 2y = 3 2x + y = 7
solving a Matrix equation In Exercises 57 and 58,
(b) x + 2y = 10 (b) 2x − y = −1 find A.
x − 2y = −6 2x + y = −3
47. (a) x1 + 2x2 + x3 = 2 57. (2A)−1 = [13 2
4 ] 58. (4A)−1 = [−32 4
2 ]
x1 + 2x2 − x3 = 4
x1 − 2x2 + x3 = −2 finding the inverse of a Matrix In Exercises 59
(b) x1 + 2x2 + x3 = 1 and 60, show that the matrix is invertible and find its
x1 + 2x2 − x3 = 3 inverse.
sin θ cos θ sec θ tan θ
x1 − 2x2 + x3 = −3
48. (a) x1 + x2 − 2x3 = 0
59. A =
−cos θ [
sin θ
60. A =
tan θ sec θ ] [ ]
x1 − 2x2 + x3 = 0 Beam deflection In Exercises 61 and 62, forces w1, w2,
x1 − x2 − x3 = −1 and w3 (in pounds) act on a simply supported elastic
(b) x1 + x2 − 2x3 = −1 beam, resulting in deflections d1, d2, and d3 (in inches) in
x1 − 2x2 + x3 = 2 the beam (see figure).
x1 − x2 − x3 = 0
d1 d2 d3
solving a system of equations Using an inverse
In Exercises 49–52, use a software program or a w1 w2 w3
graphing utility to solve the system of linear equations
using an inverse matrix.
49. x1 + 2x2 − x3 + 3x4 − x5 = −3
x1 − 3x2 + x3 + 2x4 − x5 = −3 Use the matrix equation d = Fw, where

[] []
2x1 + x2 + x3 − 3x4 + x5 = 6 d1 w1
x1 − x2 + 2x3 + x4 − x5 = 2 d= d2 , w = w2
2x1 + x2 − x3 + 2x4 + x5 = −3 d3 w3
50. x1 + x2 − x3 + 3x4 − x5 = 3 and F is the 3 × 3 flexibility matrix for the beam, to find
2x1 + x2 + x3 + x4 + x5 = 4 the stiffness matrix F −1 and the force matrix w. The
x1 + x2 − x3 + 2x4 − x5 = 3 entries of F are measured in inches per pound.
2x1 + x2 + 4x3 + x4 − x5 = −1

[ ] [ ]
0.008 0.004 0.003 0.585
3x1 + x2 + x3 − 2x4 + x5 = 5 61. F = 0.004 0.006 0.004 , d = 0.640
51. 2x1 − 3x2 + x3 − 2x4 + x5 − 4x6 = 20 0.003 0.004 0.008 0.835
3x1 + x2 − 4x3 + x4 − x5 + 2x6 = −16

[ ] [ ]
0.017 0.010 0.008 0
4x1 + x2 − 3x3 + 4x4 − x5 + 2x6 = −12 62. F = 0.010 0.012 0.010 , d = 0.15
−5x1 − x2 + 4x3 + 2x4 − 5x5 + 3x6 = −2 0.008 0.010 0.017 0
x1 + x2 − 3x3 + 4x4 − 3x5 + x6 = −15
3x1 − x2 + 2x3 − 3x4 + 2x5 − 6x6 = 25 63. proof Prove Property 2 of Theorem 2.8: If A is an
invertible matrix and k is a positive integer, then
52. 4x1 − 2x2 + 4x3 + 2x4 − 5x5 − x6 = 1
3x1 + 6x2 − 5x3 − 6x4 + 3x5 + 3x6 = −11 (Ak)−1 = A−1A−1 . . . A−1 = (A−1)k
2x1 − 3x2 + x3 + 3x4 − x5 − 2x6 = 0
−x1 + 4x2 − 4x3 − 6x4 + 2x5 + 4x6 = −9 k factors
3x1 − x2 + 5x3 + 2x4 − 3x5 − 5x6 = 1 64. proof Prove Property 4 of Theorem 2.8: If A is an
−2x1 + 3x2 − 4x3 − 6x4 + x5 + 2x6 = −12 invertible matrix, then (AT)−1 = (A−1)T.
Matrix equal to its own inverse In Exercises 53 and 65. proof Prove Property 2 of Theorem 2.10: If C is an
54, find x such that the matrix is equal to its own inverse. invertible matrix such that CA = CB, then A = B.
66. proof Prove that if A2 = A, then
53. A = [ 3
−2 −3
x
]
54. A =
−1
2
−2
x
[ ] I − 2A = (I − 2A)−1.

67. Guided Proof Prove that the inverse of a symmetric Use the result of Exercise 74 to find A−1 for each
75.
nonsingular matrix is symmetric. matrix.

[ ]
Getting Started: To prove that the inverse of A is −1 0 0
symmetric, you need to show that (A−1)T = A−1. (a) A = 0 3 0
(i) Let A be a symmetric, nonsingular matrix. 0 0 2
(ii) This means that AT = A and A−1 exists. 1

[ ]
2 0 0
1
(iii) Use the properties of the transpose to show that (b) A = 0 3 0
(A−1)T is equal to A−1. 0 0 1
4
68. Proof Prove that if A, B, and C are square matrices
and ABC = I, then B is invertible and B−1 = CA. 76. Let A = [−21 2
1 ]
.
69. Proof Prove that if A is invertible and AB = O,
(a) Show that A2 − 2A + 5I = O, where I is the
then B = O.
identity matrix of order 2.
70. Guided Proof Prove that if A2 = A, then either A is
(b) Show that A−1 = 15(2I − A).
singular or A = I.
(c) Show that for any square matrix satisfying
Getting Started: You must show that either A is singular
A2 − 2A + 5I = O, the inverse of A is
or A equals the identity matrix.
A−1 = 15(2I − A).
(i) Begin your proof by observing that A is either
singular or nonsingular. 77. Proof Let u be an n × 1 column matrix satisfying
uTu = I1. The n × n matrix H = In − 2uuT is called a
(ii) If A is singular, then you are done.
Householder matrix.
(iii) If A is nonsingular, then use the inverse matrix A−1
(a) Prove that H is symmetric and nonsingular.
and the hypothesis A2 = A to show that A = I.

[ ]
√22
True or False? In Exercises 71 and 72, determine (b) Let u = √22 . Show that uTu = I1 and find the
whether each statement is true or false. If a statement 0
is true, give a reason or cite an appropriate statement Householder matrix H.
from the text. If a statement is false, provide an example
78. Proof Let A and B be n × n matrices. Prove that if the
that shows the statement is not true in all cases or cite an
matrix I − AB is nonsingular, then so is I − BA.
appropriate statement from the text.
79. Let A, D, and P be n × n matrices satisfying AP = PD.
71. (a) If the matrices A, B, and C satisfy BA = CA and A
Assume that P is nonsingular and solve this equation
is invertible, then B = C.
for A. Must it be true that A = D?
(b) The inverse of the product of two matrices is the
80. Find an example of a singular 2 × 2 matrix satisfying
product of their inverses; that is, (AB)−1 = A−1B−1.
A2 = A.
(c) If A can be row reduced to the identity matrix, then
81. Writing Explain how to determine whether the inverse
A is nonsingular.
of a matrix exists. If so, explain how to find the inverse.
72. (a) The inverse of the inverse of a nonsingular matrix
A, (A−1)−1, is equal to A itself.
82. C
APSTONE As mentioned on page 66, if A
(b) The matrix
a
c d
b
[ ]
is invertible when ab − dc ≠ 0. is a 2 × 2 matrix
(c) If A is a square matrix, then the system of linear
equations Ax = b has a unique solution.
A= [ac b
d ]
73. Writing Is the sum of two invertible matrices then A is invertible if and only if ad − bc ≠ 0.
invertible? Explain why or why not. Illustrate your Verify that the inverse of A is
conclusion with appropriate examples.
−b
74. Writing Under what conditions will the diagonal matrix A−1 =
1 d
ad − bc −c [ a
. ]

[ ]
a11 0 0 . . . 0
0 a22 0 . . . 0
A=
⋮ ⋮ ⋮ ⋮ 83. Writing Explain in your own words how to write a
0 0 0 . . . ann system of three linear equations in three variables as a
matrix equation, Ax = b, as well as how to solve the
be invertible? Assume that A is invertible and find system using an inverse matrix.
its inverse.

2.4 Elementary Matrices

Factor a matrix into a product of elementary matrices.
Find
and use an LU-factorization of a matrix to solve a system
of linear equations.

Elementary Matrices and Elementary

Row Operations
Section 1.2 introduced the three elementary row operations for matrices listed below.
1. Interchange two rows.
2. Multiply a row by a nonzero constant.
REMARK 3. Add a multiple of a row to another row.
The identity matrix In is
elementary by this definition In this section, you will see how to use matrix multiplication to perform these operations.
because it can be obtained
from itself by multiplying any Definition of an Elementary Matrix
one of its rows by 1.
An n × n matrix is an elementary matrix when it can be obtained from the
identity matrix In by a single elementary row operation.

Elementary Matrices and

Nonelementary Matrices
Which of the matrices below are elementary? For those that are, describe the
corresponding elementary row operation.

[ ]
1 0 0
a. 0 3 0 b. [10 0
1
0
0 ]
0 0 1

[ ] [ ]
1 0 0 1 0 0
c. 0 1 0 d. 0 0 1
0 0 0 0 1 0

[ ]
1 0 0
e. [ 1
2
0
1 ] f. 0 2 0
0 0 −1

solution
a. This matrix is elementary. To obtain it from I3, multiply the second row of I3 by 3.
b. This matrix is not elementary because it is not square.
c. This matrix is not elementary because to obtain it from I3, you must multiply the
third row of I3 by 0 (row multiplication must be by a nonzero constant).
d. This matrix is elementary. To obtain it from I3, interchange the second and third
rows of I3.
e. This matrix is elementary. To obtain it from I2, multiply the first row of I2 by 2 and
add the result to the second row.
f. This matrix is not elementary because it requires two elementary row operations to
obtain from I3.

Elementary matrices are useful because they enable you to use matrix multiplication
to perform elementary row operations, as demonstrated in Example 2.

Elementary Matrices and

Elementary Row Operations
a. I n the matrix product below, E is the elementary matrix in which the first two rows
of I3 are interchanged.
E A

[ ][ ] [ ]
0 1 0 0 2 1 1 −3 6
1 0 0 1 −3 6 = 0 2 1
0 0 1 3 2 −1 3 2 −1
Note that the first two rows of A are interchanged when multiplying on the left by E.
b. In the matrix product below, E is the elementary matrix in which the second row of
I3 is multiplied by 12.
E A

[ ][ ] [ ]
1 0 0 1 0 −4 1 1 0 −4 1
1
0 2 0 0 2 6 −4 = 0 1 3 −2
0 0 1 0 1 3 1 0 1 3 1
Note that the second row of A is multiplied by 12 when multiplying on the left by E.
c. In the matrix product below, E is the elementary matrix in which 2 times the first
row of I3 is added to the second row.
E A

[ ][ ] [ ]
1 0 0 1 0 −1 1 0 −1
2 1 0 −2 −2 3 = 0 −2 1
0 0 1 0 4 5 0 4 5
Note that 2 times the first row of A is added to the second row when multiplying on
the left by E.

Notice from Example 2(b) that you can use matrix multiplication to perform
elementary row operations on nonsquare matrices. If the size of A is n × p, then E
REMARK must have order n.
Be sure to remember in In each of the three products in Example 2, you are able to perform elementary
Theorem 2.12 to multiply A row operations by multiplying on the left by an elementary matrix. The next theorem,
on the left by the elementary
stated without proof, generalizes this property of elementary matrices.
matrix E. This text does not
consider right multiplication
by elementary matrices, which
Theorem 2.12 Representing Elementary Row Operations
involves column operations.
Let E be the elementary matrix obtained by performing an elementary row
operation on Im. If that same elementary row operation is performed on an m × n
matrix A, then the resulting matrix is the product EA.

Most applications of elementary row operations require a sequence of operations.

For instance, Gaussian elimination usually requires several elementary row operations
to row reduce a matrix. This translates into multiplication on the left by several
elementary matrices. The order of multiplication is important; the elementary matrix
immediately to the left of A corresponds to the row operation performed first. Example 3
demonstrates this process.

Using Elementary Matrices

Find a sequence of elementary matrices that can be used to write the matrix A in
row-echelon form.

[ ]
0 1 3 5
A= 1 −3 0 2
2 −6 2 0

solution
Elementary Row Elementary
Matrix Operation Matrix
R1 ↔ R2

[ ] [ ]
1 −3 0 2 0 1 0
0 1 3 5 E1 = 1 0 0
2 −6 2 0 0 0 1

[ ] [ ]
1 −3 0 2 1 0 0
0 1 3 5 R3 + (−2)R1 → R3 E2 = 0 1 0
0 0 2 −4 −2 0 1

[ ] [ ]
1 −3 0 2 1 0 0
0 1 3 5 E3 = 0 1 0
0 0 1 −2 (12 )R3 → R3 0 0 1
2

The three elementary matrices E1, E2, and E3 can be used to perform the same elimination.

[ ][ ][ ][ ]
1 0 0 1 0 0 0 1 0 0 1 3 5
REMARK B = E3E2E1A = 0 1 0 0 1 0 1 0 0 1 −3 0 2
1
The procedure demonstrated 0 0 2 −2 0 1 0 0 1 2 −6 2 0
in Example 3 is primarily of

[ ][ ][ ]
theoretical interest. In other 1 0 0 1 0 0 1 −3 0 2
words, this procedure is = 0 1 0 0 1 0 0 1 3 5
not a practical method for 1
0 0 2 −2 0 1 2 −6 2 0
performing Gaussian

[ ][ ] [ ]
elimination. 1 0 0 1 −3 0 2 1 −3 0 2
= 0 1 0 0 1 3 5 = 0 1 3 5
1
0 0 2 0 0 2 −4 0 0 1 −2

The two matrices in Example 3

[ ] [ ]
0 1 3 5 1 −3 0 2
A= 1 −3 0 2 and B= 0 1 3 5
2 −6 2 0 0 0 1 −2
are row-equivalent because you can obtain B by performing a sequence of row
operations on A. That is, B = E3E2E1A.
The definition of row-equivalent matrices is restated below using elementary
matrices.

Definition of Row Equivalence

Let A and B be m × n matrices. Matrix B is row-equivalent to A when there
exists a finite number of elementary matrices E1, E2, . . . , Ek such that
B = EkEk−1 . . . E2E1A.

You know from Section 2.3 that not all square matrices are invertible. Every
elementary matrix, however, is invertible. Moreover, the inverse of an elementary
matrix is itself an elementary matrix.

Theorem 2.13 Elementary Matrices Are Invertible

If E is an elementary matrix, then E−1 exists and is an elementary matrix.

The inverse of an elementary matrix E is the elementary matrix that converts E

back to In. For instance, the inverses of the three elementary matrices in Example 3 are
shown below.
Elementary Matrix Inverse Matrix
R1 ↔ R2 R1 ↔ R2

[ ] [ ]
REMARK 0 1 0 0 1 0
E2−1 is as shown because to E1 = 1 0 0 E1−1 = 1 0 0
convert E2 back to I3, in E2 you 0 0 1 0 0 1
would add 2 times row 1 to

[ ] [ ]
row 3. 1 0 0 1 0 0
E2 = 0 1 0 E2−1 = 0 1 0
−2 0 1 R3 + (−2)R1 → R3 2 0 1 R3 + (2)R1 → R3

[ ] [ ]
1 0 0 1 0 0
E3 = 0 1 0 E3−1 = 0 1 0
0 0
1
2 (12 )R3 → R3 0 0 2 (2)R3 → R3

Use matrix multiplication to check these results.

The next theorem states that every invertible matrix can be written as the product
of elementary matrices.

Theorem 2.14 A Property of Invertible Matrices

A square matrix A is invertible if and only if it can be written as the product of
elementary matrices.

proof
The phrase “if and only if” means that there are actually two parts to the theorem.
On the one hand, you have to show that if A is invertible, then it can be written as the
product of elementary matrices. Then you have to show that if A can be written as
the product of elementary matrices, then A is invertible.
To prove the theorem in one direction, assume A is invertible. From Theorem
2.11 you know that the system of linear equations represented by Ax = O has only the
trivial solution. But this implies that the augmented matrix [A O] can be rewritten in
the form [I O] (using elementary row operations corresponding to E1, E2, . . . , and
Ek). So, Ek . . . E2E1A = I and it follows that A = E1−1E2−1 . . . Ek−1. A can be written
as the product of elementary matrices.
To prove the theorem in the other direction, assume A is the product of elementary
matrices. Every elementary matrix is invertible and the product of invertible matrices
is invertible, so it follows that A is invertible. This completes the proof.

Example 4 illustrates the first part of this proof.

Writing a Matrix as the Product

of Elementary Matrices
Find a sequence of elementary matrices whose product is the nonsingular matrix

A= [−13 −2
8
. ]
solution
Begin by finding a sequence of elementary row operations that can be used to rewrite A
in reduced row-echelon form.
Matrix Elementary Row Operation Elementary Matrix
(−1)R1 → R1
[13 2
8 ] E1 = [−10 0
1 ]
[10 2
2 ] R2 + (−3)R1 → R2
E2 = [−31 0
1 ]
[10 2
1 ] ( )R2 → R2
1
2
E3 = [10 0
1
2
]
R1 + (−2)R2 → R1 −2
[10 0
1 ] E4 = [10 1 ]
Now, from the matrix product E4E3E2E1A = I, solve for A to obtain
A = E1−1E2−1E3−1E4−1. This implies that A is a product of elementary matrices.
E1−1 E2−1 E3−1 E4−1

A= [−10 0
1 ][13 0
1][10 0
2 ][10 2
1
=
−1
] [3
−2
8 ]
In Section 2.3, you learned a process for finding the inverse of a nonsingular
matrix A. There, you used Gauss-Jordan elimination to reduce the augmented matrix
[A I] to [I A−1]. You can now use Theorem 2.14 to justify this procedure.
Specifically, the proof of Theorem 2.14 allows you to write the product
I = Ek . . . E2E1A.
Multiplying both sides of this equation (on the right) by A−1, A−1 = Ek . . . E2E1I. In
other words, a sequence of elementary matrices that reduces A to the identity I also
reduces the identity I to A−1. Applying the corresponding sequence of elementary row
operations to the matrices A and I simultaneously, you have
Ek . . . E2E1[A I] = [I A−1].
Of course, if A is singular, then no such sequence is possible.
The next theorem ties together some important relationships between n × n matrices
and systems of linear equations. The essential parts of this theorem have already been
proved (see Theorems 2.11 and 2.14); it is left to you to fill in the other parts of the proof.

Theorem 2.15 Equivalent Conditions

If A is an n × n matrix, then the statements below are equivalent.
1. A is invertible.
2. Ax = b has a unique solution for every n × 1 column matrix b.
3. Ax = O has only the trivial solution.
4. A is row-equivalent to In.
5. A can be written as the product of elementary matrices.

thE LU-factoriZation
At the heart of the most efficient and modern algorithms for solving linear systems
Ax = b is the LU-factorization, in which the square matrix A is expressed as a product,
A = LU. In this product, the square matrix L is lower triangular, which means all the
entries above the main diagonal are zero. The square matrix U is upper triangular,
which means all the entries below the main diagonal are zero.

[ ] [ ]
a11 0 0 a11 a12 a13
a21 a22 0 0 a22 a23
a31 a32 a33 0 0 a33

3 × 3 lower triangular matrix 3 × 3 upper triangular matrix

definition of LU-factorization
If the n × n matrix A can be written as the product of a lower triangular matrix
L and an upper triangular matrix U, then A = LU is an LU-factorization of A.

LU-factorizations

a. [11 2
0] [
=
1
1
0
1 ][10 2
−2
= LU]
is an LU-factorization of the matrix

A= [11 2
0 ]
as the product of the lower triangular matrix

L= [11 0
1 ]
and the upper triangular matrix

U= [10 2
−2
. ]

[ ] [ ][ ]
1 −3 0 1 0 0 1 −3 0
b. A = 0 1 3 = 0 1 0 0 1 3 = LU
2 −10 2 2 −4 1 0 0 14

is an LU-factorization of the matrix A.

Computational fluid dynamics (CFD) is the computer-based

linEar simulation of such real-life phenomena as fluid flow, heat
alGEBra transfer, and chemical reactions. Solving the conservation
appliEd of energy, mass, and momentum equations involved in a
CFD analysis can involve large systems of linear equations.
So, for efficiency in computing, CFD analyses often use
matrix partitioning and LU-factorization in their algorithms.
Aerospace companies such as Boeing and Airbus have
used CFD analysis in aircraft design. For instance,
engineers at Boeing used CFD analysis to simulate
airflow around a virtual model of their 787 aircraft to help
produce a faster and more efficient design than those of
earlier Boeing aircraft.
Goncharuk/Shutterstock.com

If a square matrix A row reduces to an upper triangular matrix U using only

the row operation of adding a multiple of one row to another row below it, then it is
relatively easy to find an LU-factorization of the matrix A. All you need to do is keep
track of the individual row operations, as shown in the next example.

Finding an LU-Factorization of a Matrix

[ ]
1 −3 0
Find an LU-factorization of the matrix A = 0 1 3 .
2 −10 2

solution
Begin by row reducing A to upper triangular form while keeping track of the elementary
matrices used for each row operation.
Matrix Elementary Row Operation Elementary Matrix

[ ] [ ]
1 −3 0 1 0 0
0 1 3 E1 = 0 1 0
0 −4 2 R3 + (−2)R1 → R3 −2 0 1

[ ] [ ]
1 −3 0 1 0 0
0 1 3 E2 = 0 1 0
0 0 14 R3 + (4)R2 → R3 0 4 1

The reduced matrix above is an upper triangular matrix U, and it follows that
E2E1A = U, or A = E1−1E2−1U. The product of the lower triangular matrices

[ ][ ] [ ]
1 0 0 1 0 0 1 0 0
E1−1E2−1 = 0 1 0 0 1 0 = 0 1 0
2 0 1 0 −4 1 2 −4 1
is a lower triangular matrix L, so the factorization A = LU is complete. Notice that this
is the same LU-factorization as in Example 5(b).

If A row reduces to an upper triangular matrix U using only the row operation of
adding a multiple of one row to another row below it, then A has an LU-factorization.
Ek . . . E2E1A = U
A = E1−1E2−1 . . . Ek−1U = LU
Here L is the product of the inverses of the elementary matrices used in the row reduction.
Note that the multipliers in Example 6 are −2 and 4, which are the negatives of
the corresponding entries in L. This is true in general. If U can be obtained from A
using only the elementary row operation of adding a multiple of one row to another
row below it, then the matrix L is lower triangular (with 1’s along the diagonal), and
the negative of each multiplier is in the same position as that of the corresponding zero
in U below the main diagonal.
Once you have obtained an LU-factorization of a matrix A, you can then solve the
system of n linear equations in n variables Ax = b very efficiently in two steps.
1. Write y = Ux and solve Ly = b for y.
2. Solve Ux = y for x.
The column matrix x is the solution of the original system because Ax = LUx = Ly = b.
The second step is just back-substitution, because the matrix U is upper triangular.
The first step is similar, except that it starts at the top of the matrix, because L is lower
triangular. For this reason, the first step is often called forward substitution.

solving a linear system using

LU-factorization
See LarsonLinearAlgebra.com for an interactive version of this type of example.

Solve the linear system.

x1 − 3x2 = −5
x2 + 3x3 = −1
2x1 − 10x2 + 2x3 = −20
solution
You obtained an LU-factorization of the coefficient matrix A in Example 6.

[ ]
1 −3 0
A= 0 1 3
2 −10 2

[ ][ ]
1 0 0 1 −3 0
= 0 1 0 0 1 3
2 −4 1 0 0 14
First, let y = Ux and solve the system Ly = b for y.

[ ][ ] [ ]
1 0 0 y1 −5
0 1 0 y2 = −1
2 −4 1 y3 −20
Solve this system using forward substitution. Starting with the first equation, you have
y1 = −5. The second equation gives y2 = −1. Finally, from the third equation,
2y1 − 4y2 + y3 = −20
y3 = −20 − 2y1 + 4y2
y3 = −20 − 2(−5) + 4(−1)
y3 = −14.
The solution of Ly = b is

[ ]
−5
y= −1 .
−14
Now solve the system Ux = y for x using back-substitution.

[ ][ ] [ ]
1 −3 0 x1 −5
0 1 3 x2 = −1
0 0 14 x3 −14
From the bottom equation, x3 = −1. Then, the second equation gives
x2 + 3(−1) = −1
or x2 = 2. Finally, the first equation gives
x1 − 3(2) = −5
or x1 = 1. So, the solution of the original system of equations is

[ ]
1
x= 2 .
−1

2.4 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

Elementary Matrices In Exercises 1–8, determine Finding the Inverse of an Elementary Matrix
whether the matrix is elementary. If it is, state the In Exercises 19–24, find the inverse of the elementary
elementary row operation used to produce it. matrix.
1.
1
0[ 0
2
2. ] 1
0
0
0
0
1 [ ] 19.
0
1 [1
0 ]
20.
5
0
0
1 [ ]
1
[ 0
] 0 1
[ ]
[ ] [ ]
3. 4. 0 0 1 1 0 0
2 1 1 0 21. 0 1 0 22. 0 1 0

[ ] [ ]
2 0 0 1 0 0 1 0 0 0 −3 1
5. 0 0 1 6. 0 1 0

[ ]
0 1 0 2 0 1 1 0 0 0

[ ]
k 0 0
0 1 k 0

[ ]
1 0 0 0 23. 0 1 0 24.
0 0 1 0
0 1 0 0 0 0 1
7. 0 0 0 1
0 −5 1 0 k≠0
0 0 0 1
Finding the Inverse of a Matrix In Exercises 25–28,

[ ]
1 0 0 0 find the inverse of the matrix using elementary matrices.
2 1 0 0 3 −2
8.
0 0 1 0 25.
1 [ 0 ]26.
2
1
0
1 [ ]
0 0 −3 1

[ ] [ ]
1 0 −1 1 0 −2
Finding an Elementary Matrix In Exercises 9–12, 27. 0 6 −1 28. 0 2 1
let A, B, and C be 0 0 4 0 0 1

[ ] [ ]
1 2 −3 −1 2 0 Finding a Sequence of Elementary Matrices
A= 0 1 2 , B= 0 1 2 , and In Exercises 29–36, find a sequence of elementary
−1 2 0 1 2 −3 matrices whose product is the given nonsingular matrix.

[ ] [ ] [ ]
0 4 −3 1 2 0 1
29. 30.
C= 0 1 2 . 1 0 1 0
−1 −1
31. [
−1]
32. [
1]
2 0 4 1 1
9. Find an elementary matrix E such that EA = B. 3 2

[ ] [ ]
10. Find an elementary matrix E such that EA = C. 1 −2 0 1 2 3
33. −1 3 0 34. 2 5 6
11. Find an elementary matrix E such that EB = A.
0 0 1 1 3 4
12. Find an elementary matrix E such that EC = A.

[ ] [ ]
1 0 0 1 4 0 0 2
Finding a Sequence of Elementary Matrices 0 −1 3 0 0 1 0 1
In Exercises 13–18, find a sequence of elementary matrices 35. 36.
0 0 2 0 0 0 −1 2
that can be used to write the matrix in row-echelon form.
0 0 1 −1 1 0 0 −2

[ ]
0 3 −3 6
13. [
0 1
5 10 −5
7
]
14. 1 −1 2 −2 37. Writing Is the product of two elementary matrices
0 0 2 2 always elementary? Explain.

[ ] [ ]
1 −2 −1 0 1 3 0 38. Writing E is the elementary matrix obtained by
15. 0 4 8 −4 16. 2 5 −1 interchanging two rows in In. A is an n × n matrix.
−6 12 8 1 3 −2 −4 (a) How will EA compare with A? (b) Find E 2.

[ ] [ ]
−2 1 0 1 −6 0 2 39. Use elementary matrices to find the inverse of
3 −4 0 0 −3 3 9

[ ]
17. 18. 1 0 0
1 −2 2 2 5 −1 1
A= 0 1 0 , c ≠ 0.
−1 2 −2 4 8 −5 1
a b c

40. Use elementary matrices to find the inverse of 53. Determine a and b such that A is idempotent.

[ ][ ][ ]
1 a 0 1 0 0 1 0 0
A= 0 1 0 b 1 0 0 1 0 , A= [1a 0
b ]
0 0 1 0 0 1 0 0 c
54. Guided Proof Prove that A is idempotent if and only
c ≠ 0. if AT is idempotent.
True or False? In Exercises 41 and 42, determine Getting Started: The phrase “if and only if” means that
whether each statement is true or false. If a statement you have to prove two statements:
is true, give a reason or cite an appropriate statement 1. If A is idempotent, then AT is idempotent.
from the text. If a statement is false, provide an example 2. If AT is idempotent, then A is idempotent.
that shows the statement is not true in all cases or cite an
(i) Begin your proof of the first statement by assuming
appropriate statement from the text.
that A is idempotent.
41. (a) The identity matrix is an elementary matrix.
(ii) This means that A2 = A.
(b) If E is an elementary matrix, then 2E is an
(iii) Use the properties of the transpose to show that AT
elementary matrix.
is idempotent.
(c) The inverse of an elementary matrix is an
(iv) Begin your proof of the second statement by
elementary matrix.
assuming that AT is idempotent.
42. (a) The zero matrix is an elementary matrix.
55. Proof Prove that if A is an n × n matrix that is
(b)
A square matrix is nonsingular when it can be idempotent and invertible, then A = In.
written as the product of elementary matrices.
56. Proof Prove that if A and B are idempotent and
(c) Ax = O has only the trivial solution if and only AB = BA, then AB is idempotent.
if Ax = b has a unique solution for every n × 1
57. Guided Proof Prove that if A is row-equivalent to B
column matrix b.
and B is row-equivalent to C, then A is row-equivalent
Finding an LU-Factorization of a Matrix In to C.
Exercises 43–46, find an LU-factorization of the matrix. Getting Started: To prove that A is row-equivalent to C,
−2 you have to find elementary matrices E1, E2, . . . , Ek
43.
−2
1
[ 0
1
44. ]−6
1
4 [ ] such that A = Ek . . . E2 E1C.
(i) Begin by observing that A is row-equivalent to B

[ ] [ ]
3 0 1 2 0 0
45. 6 1 1 46. 0 −3 1 and B is row-equivalent to C.
−3 1 0 10 12 3 (ii) This means that there exist elementary matrices
F1, F2, . . . , Fn and G1, G2, . . . , Gm such that
Solving a Linear System Using LU-Factorization A = Fn . . . F2F1B and B = Gm . . . G2G1C.
In Exercises 47 and 48, use an LU-factorization of the (iii) Combine the matrix equations from step (ii).
coefficient matrix to solve the linear system.
58. Proof Prove that if A is row-equivalent to B, then B is
47. 2x + y = 1 row-equivalent to A.
y−z= 2 59. Proof Let A be a nonsingular matrix. Prove that if B
−2x + y + z = −2 is row-equivalent to A, then B is also nonsingular.
48. 2x1 = 4
−2x1 + x2 − x3 = −4
60. CAPSTONE
6x1 + 2x2 + x3 = 15
(a) Explain how to find an elementary matrix.
−x4 = −1
(b) Explain how to use elementary matrices to find an
Idempotent Matrices In Exercises 49–52, determine LU-factorization of a matrix.
whether the matrix is idempotent. A square matrix A is (c) Explain how to use LU-factorization to solve a
idempotent when A2 = A. linear system.
49.
1
0 [ 0
0 ]
50.
0
1
1
0 [ ]
61.
Show that the matrix below does not have an

[ ] [ ]
0 0 1 0 1 0 LU-factorization.
51. 0 1 0 52. 1 0 0
1 0 0 0 0 1 A= [01 1
0 ]

2.5 Markov Chains

Use a stochastic matrix to find the nth state matrix of a Markov
chain.
Find the steady state matrix of a Markov chain.
Find the steady state matrix of an absorbing Markov chain.

Stochastic Matrices and markov chains

Many types of applications involve a finite set of states { S1, S2, . . . , Sn } of a
population. For instance, residents of a city may live downtown or in the suburbs.
Voters may vote Democrat, Republican, or Independent. Soft drink consumers may buy
Coca-Cola, Pepsi Cola, or another brand.
The probability that a member of a population will change from the jth state to the
ith state is represented by a number pij, where 0 ≤ pij ≤ 1. A probability of pij = 0
means that the member is certain not to change from the jth state to the ith state,
whereas a probability of pij = 1 means that the member is certain to change from the
jth state to the ith state.
From

S1 S2 . . . Sn

[ ]}
p11 p12 . . . p1n S1
p21 p22 . . . p2n S2
P= To
⋮ ⋮ ⋮ ⋮
pn1 pn2 . . . pnn Sn
P is called the matrix of transition probabilities because it gives the probabilities of
each possible type of transition (or change) within the population.
At each transition, each member in a given state must either stay in that state or
change to another state. For probabilities, this means that the sum of the entries in any
column of P is 1. For instance, in the first column,
p11 + p21 + . . . + pn1 = 1.
Such a matrix is called stochastic (the term “stochastic” means “regarding
conjecture”). That is, an n × n matrix P is a stochastic matrix when each entry is a
number between 0 and 1 inclusive, and the sum of the entries in each column of P is 1.

Examples of Stochastic Matrices

and Nonstochastic Matrices
The matrices in parts (a), (b), and (c) are stochastic, but the matrices in parts (d), (e),
and (f) are not.

[ ]
1 1 1 1
4 5 3 2

[ ]
1 0 0 1 13
0
1
a. 0 1 0 b.
4
1
60
1 1
6
1 c. [0.9
0.1
0.8
0.2 ]
0 0 1 4 3 3 6
1 1 1 1
4 4 3 6

[ ] [ ]
1
2
1
4
1
4
0.1 0.2 0.3 0.4

[ ]
1 1
2 4 1 2 0.2 0.3 0.4 0.5
d. 3 e. 3 0 3 f.
0 4
0.3 0.4 0.5 0.6
1 3
4 4 0 0.4 0.5 0.6 0.7

Example 2 describes the use of a stochastic matrix to measure consumer preferences.

A Consumer Preference Model

Two competing companies offer satellite television service to a city with 100,000
households. Figure 2.1 shows the changes in satellite subscriptions each year.
20% Company A now has 15,000 subscribers and Company B has 20,000 subscribers. How
Company Company many subscribers will each company have in one year?
A B
solution
15%
70% 80%
The matrix of transition probabilities is
15% 15%
10% From
5%

No A B None
Satellite

[ ]
0.70 0.15 0.15 A
Television
P = 0.20 0.80 0.15 B To
0.10 0.05 0.70 None
70%
and the initial state matrix representing the portions of the total population in the three
Figure 2.1
states is

[ ]
0.1500 A
X0 = 0.2000 . B
0.6500 None

To find the state matrix representing the portions of the population in the three states in
one year, multiply P by X0 to obtain

[ ][ ] [ ]
0.70 0.15 0.15 0.1500 0.2325
X1 = PX0 = 0.20 0.80 0.15 0.2000 = 0.2875 .
0.10 0.05 0.70 0.6500 0.4800

REMARK In one year, Company A will have 0.2325(100,000) = 23,250 subscribers and
Always assume that the matrix Company B will have 0.2875(100,000) = 28,750 subscribers.
P of transition probabilities in a
Markov chain remains constant A Markov chain, named after Russian mathematician Andrey Andreyevich Markov
between states. (1856–1922), is a sequence { Xn } of state matrices that are related by the equation
Xk+1 = PXk, where P is a stochastic matrix. For instance, consider the consumer
preference model discussed in Example 2. To find the state matrix representing the
portions of the population in each state in three years, repeatedly multiply the initial
state matrix X0 by the matrix of transition probabilities P.
X1 = PX0
X2 = PX1 = P ∙ PX0 = P2X0
X3 = PX2 = P ∙ P2X0 = P3X0
In general, the nth state matrix of a Markov chain is PnX0, as summarized below.

nth State Matrix of a Markov Chain

The nth state matrix of a Markov chain for which P is the matrix of transition
probabilities and X0 is the initial state matrix is
Xn = PnX0.

Example 3 uses the model discussed in Example 2 to demonstrate this process.

a consumer Preference Model

Assuming the matrix of transition probabilities from Example 2 remains the same year
after year, find the number of subscribers each satellite television company will have after
(a) 3 years, (b) 5 years, (c) 10 years, and (d) 15 years.
Solution
a. To find the numbers of subscribers after 3 years, first find X3.

[ ]
0.3028 A
X3 = P3X0 ≈ 0.3904 B After 3 years
0.3068 None
After 3 years, Company A will have about 0.3028(100,000) = 30,280 subscribers
and Company B will have about 0.3904(100,000) = 39,040 subscribers.
b. To find the numbers of subscribers after 5 years, first find X5.

[ ]
0.3241 A
X5 = P5X0 ≈ 0.4381 B After 5 years
0.2378 None
After 5 years, Company A will have about 0.3241(100,000) = 32,410 subscribers
and Company B will have about 0.4381(100,000) = 43,810 subscribers.
c. To find the numbers of subscribers after 10 years, first find X10.

[ ]
0.3329 A
X10 = P10X0 ≈ 0.4715 B After 10 years
0.1957 None
After 10 years, Company A will have about 0.3329(100,000) = 33,290 subscribers
and Company B will have about 0.4715(100,000) = 47,150 subscribers.
d. To find the numbers of subscribers after 15 years, first find X15.

[ ]
0.3333 A
X15 = P15X0 ≈ 0.4756 B After 15 years
0.1911 None
After 15 years, Company A will have about 0.3333(100,000) = 33,330 subscribers
and Company B will have about 0.4756(100,000) = 47,560 subscribers.

Google’s PageRank algorithm makes use of Markov chains.

linear For a search set that contains n web pages, define an n × n
algebra matrix A such that aij = 1 when page j references page i
aPPlied and aij = 0 otherwise. Adjust A to account for web pages
without external references, scale each column of A so
that A is stochastic, and call this matrix B. Then define
1−p
M = pB + E
n
where p is the probability that a user follows a link on a
page, 1 − p is the probability that the user goes to any
page at random, and E is an n × n matrix whose entries
are all 1. The Markov chain whose matrix of transition
probabilities is M converges to a unique steady state
matrix, which gives an estimate of page ranks.
Section 10.3 discusses a method that can be used to
estimate the steady state matrix.
d8nn/Shutterstock.com

Steady state matrix of a markov chain

In Example 3, notice that there is little difference between the numbers of subscribers
after 10 years and after 15 years. If you continue the process shown in this example,
then the state matrix Xn eventually reaches a steady state. That is, as long as the matrix
P does not change, the matrix product PnX approaches a limit X. In Example 3, the limit
is the steady state matrix

[]
1

[ ]
3 0.3333 A
10
X= 21 ≈ 0.4762 . B Steady state matrix
4 0.1905 None
21

Check to see that PX = X, as shown below.

][ ] []
1 1

[
0.70 0.15 0.15 3 3
10 10
PX = 0.20 0.80 0.15 21 = 21 =X
0.10 0.05 0.70 4 4
21 21

In Example 5, you will verify the above result by finding the steady state matrix X.
The matrix of transition probabilities P used above is an example of a regular
stochastic matrix. A stochastic matrix P is regular when some power of P has only
positive entries.

Regular Stochastic Matrices

a. The stochastic matrix

[ ]
0.70 0.15 0.15
P = 0.20 0.80 0.15
0.10 0.05 0.70
is regular because P1 has only positive entries.
b. The stochastic matrix

P= [0.50
0.50
1.00
0 ]
is regular because

P2 = [0.75
0.25
0.50
0.50 ]
has only positive entries.
c. The stochastic matrix
REMARK

[ ]
1
For a regular stochastic matrix 3 0 1
P, the sequence of successive P=
1
1 0
3
powers 1
3 0 0
P, P 2, P 3, . . .
is not regular because every power of P has two zeros in its second column.
approaches a stable matrix P. (Verify this.)
The entries in each column of
P are equal to the corresponding
entries in the steady state When P is a regular stochastic matrix, the corresponding regular Markov chain
matrix X . You are asked to PX0, P2X0, P3X0, . . .
show this in Exercise 55.
approaches a unique steady state matrix X. You are asked to prove this in Exercise 56.

Finding a Steady State Matrix

See LarsonLinearAlgebra.com for an interactive version of this type of example.

Find the steady state matrix X of the Markov chain whose matrix of transition
probabilities is the regular matrix

[ ]
0.70 0.15 0.15
P = 0.20 0.80 0.15 .
0.10 0.05 0.70

Solution
Note that P is the matrix of transition probabilities that you found in Example 2 and
whose steady state matrix X you verified at the top of page 87. To find X, begin by

[]
x1
letting X = x2 . Then use the matrix equation PX = X to obtain
x3

[ ][ ] [ ]
0.70 0.15 0.15 x1 x1
0.20 0.80 0.15 x2 = x2
reMark 0.10 0.05 0.70 x3 x3
Recall from Example 2 that the or
state matrix consists of entries
that are portions of the whole. 0.70x1 + 0.15x2 + 0.15x3 = x1
So it should make sense that 0.20x1 + 0.80x2 + 0.15x3 = x2
x1 + x2 + x3 = 1. 0.10x1 + 0.05x2 + 0.70x3 = x3.
Use these equations and the fact that x1 + x2 + x3 = 1 to write the system of linear
equations below.
−0.30x1 + 0.15x2 + 0.15x3 =0
0.20x1 − 0.20x2 + 0.15x3 =0
0.10x1 + 0.05x2 − 0.30x3 =0
x1 + x2 + x3 = 1.
Use any appropriate method to verify that the solution of this system is
x1 = 13, x2 = 10 4
21 , and x3 = 21 .
So the steady state matrix is

[]
1

[ ]
3 0.3333
10
X= 21 ≈ 0.4762 .
4 0.1905
21

Check that PX = X.
reMark
If P is not regular, then the A summary for finding the steady state matrix X of a Markov chain is below.
corresponding Markov chain
may or may not have a unique
Finding the Steady State Matrix of a Markov chain
steady state matrix.
1. Check to see that the matrix of transition probabilities P is a regular matrix.
2. Solve the system of linear equations obtained from the matrix equation
PX = X along with the equation x1 + x2 + . . . + xn = 1.
3. Check the solution found in Step 2 in the matrix equation PX = X.

Absorbing markov chains

The Markov chain discussed in Examples 3 and 5 is regular. Other types of Markov
REMARK chains can be used to model real-life situations. One of these includes absorbing
In Exercise 50, you will Markov chains.
investigate another type of Consider a Markov chain with n different states { S1, S2, . . . , Sn }. The ith state Si
Markov chain, one with is an absorbing state when, in the matrix of transition probabilities P, pii = 1. That
reflecting boundaries. is, the entry on the main diagonal of P is 1 and all other entries in the ith column of P
are 0. An absorbing Markov chain has the two properties listed below.
1. The Markov chain has at least one absorbing state.

2. It is possible for a member of the population to move from any nonabsorbing state
to an absorbing state in a finite number of transitions.

40% S1 Absorbing and Nonabsorbing Markov Chains

60%
a. For the matrix
From
S2 S3

50% S1 S2 S3

[ ]
100% 50% 0.4 0 0 S1
Figure 2.2 P= 0 1 0.5 S2 To
0.6 0 0.5 S3
t he second state, represented by the second column, is absorbing. Moreover, the
S1 S2
corresponding Markov chain is also absorbing because it is possible to move from
S1 to S2 in two transitions, and it is possible to move from S3 to S2 in one transition.
50%
(See Figure 2.2.)
50% 100% b. For the matrix
60%
From
S3 S4
S1 S2 S3 S4

[ ]
50% 0.5 0 0 0 S1
40% 50%
0.5 1 0 0 S2
Figure 2.3 P= To
0 0 0.4 0.5 S3
0 0 0.6 0.5 S4
the second state is absorbing. However, the corresponding Markov chain is not
S4 S2 absorbing because there is no way to move from state S3 or state S4 to state S2. (See
Figure 2.3.)
20% 20%
100% 100 % c. The matrix
S1 From
10% 30%
50% 20% S1 S2 S3 S4
10%

[ ]
0.5 0 0.2 0 S1
S3 0.2 1 0.3 0 S2
P= To
0.1 0 0.4 0 S3
0.2 0 0.1 1 S4
40%
Figure 2.4
h as two absorbing states: S2 and S4. Moreover, the corresponding Markov chain is
also absorbing because it is possible to move from either of the nonabsorbing
states, S1 or S3, to either of the absorbing states in one step. (See Figure 2.4.)

It is possible for some absorbing Markov chains to have a unique steady state
matrix. Other absorbing Markov chains have an infinite number of steady state
matrices. Example 7 demonstrates.

Finding Steady State Matrices of Absorbing

Markov Chains

Find the steady state matrix X of each absorbing Markov chain with matrix of transition
probabilities P.

[ ]
0.5 0 0.2 0

[ ]
0.4 0 0
0.2 1 0.3 0
a. P = 0 1 0.5 b. P =
0.1 0 0.4 0
0.6 0 0.5
0.2 0 0.1 1

solution
a. Use the matrix equation PX = X, or

[ ][ ] [ ]
0.4 0 0 x1 x1
0 1 0.5 x2 = x2
0.6 0 0.5 x3 x3
along with the equation x1 + x2 + x3 = 1 to write the system of linear equations
−0.6x1 =0
0.5x3 = 0
0.6x1 − 0.5x3 = 0
x1 + x2 + x3 = 1.
The solution of this system is x1 = 0, x2 = 1, and x3 = 0, so the steady state matrix
is X = [0 1 0]T. Note that X coincides with the second column of the matrix of
transition probabilities P.
b. Use the matrix equation PX = X, or

[ ][ ] [ ]
0.5 0 0.2 0 x1 x1
0.2 1 0.3 0 x2 x
= 2
0.1 0 0.4 0 x3 x3
0.2 0 0.1 1 x4 x4
along with the equation x1 + x2 + x3 + x4 = 1 to write the system of linear equations

REMARK −0.5x1 + 0.2x3 =0

Note that the steady state 0.2x1 + 0.3x3 =0
matrix for an absorbing 0.1x1 − 0.6x3 =0
Markov chain has nonzero 0.2x1 + 0.1x3 =0
values only in the absorbing x1 + x2 + x3 + x4 = 1.
state(s). These states absorb
the population. he solution of this system is x1 = 0, x2 = 1 − t, x3 = 0, and x4 = t, where t is any real
T
number such that 0 ≤ t ≤ 1. So, the steady state matrix is X = [0 1 − t 0 t]T.
The Markov chain has an infinite number of steady state matrices.

In general, a regular Markov chain or an absorbing Markov chain with one

absorbing state has a unique steady state matrix regardless of the initial state matrix.
Further, an absorbing Markov chain with two or more absorbing states has an infinite
number of steady state matrices, which depend on the initial state matrix. In Exercise
49, you are asked to show this dependence for the Markov chain whose matrix of
transition probabilities is given in Example 7(b).

2.5 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

Stochastic Matrices In Exercises 1–6, determine 11. Purchase of a Product The market research
whether the matrix is stochastic. department at a manufacturing plant determines that
20% of the people who purchase the plant’s product
[ ]
2
− 25
1.
5
3
5
7
5
2. [1 +− √
√2
2 1 − √2
√2 ] during any month will not purchase it the next month.
On the other hand, 30% of the people who do not
purchase the product during any month will purchase

[ ] [ ]
0.3 0.16 0.25 0.3 0.5 0.2
it the next month. In a population of 1000 people,
3. 0.3 0.6 0.25 4. 0.1 0.2 0.7
100 people purchased the product this month. How
0.3 0.16 0.5 0.8 0.1 0.1 many will purchase the product (a) next month and (b) in
2 months?

[ ]
1 0 0 0
0 1 0 0 12. Spread of a Virus A medical researcher is studying
5. the spread of a virus in a population of 1000 laboratory
0 0 1 0
0 0 0 1 mice. During any week, there is an 80% probability
that an infected mouse will overcome the virus, and

[ ]
1 2 1 4 during the same week there is a 10% probability that a
2 9 4 15
1 1 1 4
noninfected mouse will become infected. Three
6 3 4 15 hundred mice are currently infected with the virus. How
6. 1 2 1 4 many will be infected (a) next week and (b) in 3 weeks?
6 9 4 15
1 2 1 1 13. Television Watching A college dormitory houses
6 9 4 5
200 students. Those who watch an hour or more of
television on any day always watch for less than an hour
7. Airplane Allocation An airline has 30 airplanes in
the next day. One-fourth of those who watch television
Los Angeles, 12 airplanes in St. Louis, and 8 airplanes
for less than an hour one day will watch an hour or more
in Dallas. During an eight-hour period, 20% of the
the next day. Half of the students watched television for
planes in Los Angeles fly to St. Louis and 10% fly to
an hour or more today. How many will watch television
Dallas. Of the planes in St. Louis, 25% fly to Los Angeles
for an hour or more (a) tomorrow, (b) in 2 days, and (c)
and 50% fly to Dallas. Of the planes in Dallas, 12.5%
in 30 days?
fly to Los Angeles and 50% fly to St. Louis. How many
planes are in each city after 8 hours? 14. Sports Activities Students in a gym class have a
choice of swimming or playing basketball each day.
8. Chemistry In a chemistry experiment, a test tube
Thirty percent of the students who swim one day will
contains 10,000 molecules of a compound. Initially, 20%
swim the next day. Sixty percent of the students who
of the molecules are in a gas state, 60% are in a liquid
play basketball one day will play basketball the next
state, and 20% are in a solid state. After introducing
day. Today, 100 students swam and 150 students played
a catalyst, 40% of the gas molecules change to liquid,
basketball. How many students will swim (a) tomorrow,
30% of the liquid molecules change to solid, and 50%
(b) in two days, and (c) in four days?
of the solid molecules change to liquid. How many
molecules are in each state after introducing the catalyst? 15. Smokers and Nonsmokers In a population of
10,000, there are 5000 nonsmokers, 2500 smokers of
Finding State Matrices In Exercises 9 and 10, use one pack or less per day, and 2500 smokers of more
the matrix of transition probabilities P and initial state than one pack per day. During any month, there is a
matrix X0 to find the state matrices X1, X2, and X3. 5% probability that a nonsmoker will begin smoking
a pack or less per day, and a 2% probability that a

[ ] [ ]
0.6 0.1 0.1 0.1 nonsmoker will begin smoking more than a pack
9. P = 0.2 0.7 0.1 , X0 = 0.1 per day. For smokers who smoke a pack or less per
0.2 0.2 0.8 0.8 day, there is a 10% probability of quitting and a 10%
probability of increasing to more than a pack per day.

[]
1

[ ]
0.6 0.2 0 3 For smokers who smoke more than a pack per day, there
1
10. P = 0.2 0.7 0.1 , X0 = 3 is a 5% probability of quitting and a 10% probability of
0.2 0.1 0.9 1 dropping to a pack or less per day. How many people
3
will be in each group (a) in 1 month, (b) in 2 months,
and (c) in 1 year?

16. Consumer Preference In a population of 100,000 31. (a) Find the steady state matrix X using the matrix of
consumers, there are 20,000 users of Brand A, 30,000 transition probabilities P in Exercise 9.
users of Brand B, and 50,000 who use neither brand. (b) Find the steady state matrix X using the matrix of
During any month, a Brand A user has a 20% probability transition probabilities P in Exercise 10.
of switching to Brand B and a 5% probability of not
using either brand. A Brand B user has a 15% probability 32. Find the steady state matrix for each stochastic matrix
of switching to Brand A and a 10% probability of not in Exercises 1–6.
using either brand. A nonuser has a 10% probability of 33. Fundraising A nonprofit organization collects
purchasing Brand A and a 15% probability of purchasing contributions from members of a community. During
Brand B. How many people will be in each group (a) in any year, 40% of those who make contributions will
1 month, (b) in 2 months, and (c) in 18 months? not contribute the next year. On the other hand, 10% of
those who do not make contributions will contribute the
Regular and Steady State Matrices In Exercises next year. Find and interpret the steady state matrix for
17–30, determine whether the stochastic matrix P is this situation.
regular. Then find the steady state matrix X of the 34. Grade Distribution In a college class, 70% of the
Markov chain with matrix of transition probabilities P. students who receive an “A” on one assignment will
receive an “A” on the next assignment. On the other
17. P = [0.5
0.5
0.1
0.9 ] 18. P = [01 0.3
0.7 ] hand, 10% of the students who do not receive an “A”
on one assignment will receive an “A” on the next
19. P = [10 0.75
0.25 ] 20. P = [0.2
0.8
0
1] assignment. Find and interpret the steady state matrix
for this situation.

[ ] [ ]
1 1 2 7
2 3 5 10 35. Stock Sales and Purchases Eight hundred fifty
21. P = 1 2 22. P = 3 3 stockholders invest in one of three stocks. During any
2 3 5 10
month, 25% of Stock A holders move their investment

[ ] [ ]
2 3 1 2 1 1
5 10 2 9 4 3 to Stock B and 10% to Stock C. Of Stock B holders,
23. P =
1 1 1
24. P =
1 1 1 10% move their investment to Stock A. Of Stock C
5 5 10 3 2 3
2 1 2 4 1 1
holders, 15% move their investment to Stock A and 5%
5 2 5 9 4 3 to Stock B. Find and interpret the steady state matrix for
this situation.

[ ]
1 0 0.15
25. P = 0 1 0.10 36. Customer Preference Two movie theatres that
0 0 0.75 show several different movies each night compete for
the same audience. Of the people who attend Theatre A

[ ]
1 1
2 5 1 one night, 10% will attend again the next night and 5%
26. P =
1 1
0 will attend Theatre B the next night. Of the people who
3 5
1 3 attend Theatre B one night, 8% will attend again the
6 5 0 next night and 6% will attend Theatre A the next night.

[ ]
0.22 0.20 0.65 Of the people who attend neither theatre one night, 3%
27. P = 0.62 0.60 0.15 will attend Theatre A the next night and 4% will attend
0.16 0.20 0.20 Theatre B the next night. Find and interpret the steady
state matrix for this situation.

[ ]
0.1 0 0.3
28. P = 0.7 1 0.3 Absorbing Markov Chains In Exercises 37–40,
0.2 0 0.4 determine whether the Markov chain with matrix of
transition probabilities P is absorbing. Explain.

[ ]
1 1 1
1

[ ] [ ]
4 3 2 0.8 0.3 0 1 0 0
1 1 1
4 3 2 0 37. P = 0.2 0.1 0 38. P = 0 0.3 0.9
29. P = 1 1 0 0.6 1 0 0.7 0.1
4 3 0 0

[ ]
1 2 1
0 0 0 0 0

[ ]
4 5 5
0.3 0.7 0.2 0
1 3 1
0

[ ]
1 0 0 0 5 5 2 0.2 0.1 0.1 0
39. P = 2 1 40. P =
0 0 1 0 5 5 1 0 0.1 0.1 0.1 0
30. P = 0.4 0.1 0.6 1
0 1 0 0 0 0 0 1
2
0 0 0 1

Finding a Steady State Matrix In Exercises 41–44, 50. Markov Chain with Reflecting Boundaries The
find the steady state matrix X of the absorbing Markov figure below illustrates an example of a Markov chain
chain with matrix of transition probabilities P. with reflecting boundaries.

[ ] [ ]
0.6 0 0.3 0.1 0 0 100% 60% 70%
41. P = 0.2 1 0.6 42. P = 0.2 1 0
S1 S2 S3 S4
0.2 0 0.1 0.7 0 1
40% 30% 100%

[ ]
1 0.2 0.1 0.3
0 0.3 0.6 0.3 (a) Explain why it is appropriate to say that this type of
43. P =
0 0.1 0.2 0.2 Markov chain has reflecting boundaries.
0 0.4 0.1 0.2 (b) Use the figure to write the matrix of transition

[ ]
0.7 0 0.2 0.1 probabilities P for the Markov chain.
0.1 1 0.5 0.6 (c) Find P30 and P31. Find several other high even
44. P = powers 2n and odd powers 2n + 1 of P. What do
0 0 0.2 0.2
0.2 0 0.1 0.1 you observe?
(d) Find the steady state matrix X of the Markov chain.
45. Epidemic Model In a population of 200,000 people, How are the entries in the columns of P2n and P2n+1
40,000 are infected with a virus. After a person becomes related to the entries in X?
infected and then recovers, the person is immune
(cannot become infected again). Of the people who Nonabsorbing Markov Chain In Exercises 51 and 52,
are infected, 5% will die each year and the others will consider the matrix P in Example 6(b).
recover. Of the people who have never been infected, 51. Is it possible to find a steady state matrix X for the
25% will become infected each year. How many people corresponding Markov chain? If so, find a steady state
will be infected in 4 years? matrix. If not, explain why.
46. Chess Tournament Two people are engaged in a 52. Create a new matrix P′ by changing the second column
chess tournament. Each starts with two playing chips. of P to [0.6 0.4 0 0]T, resulting in a second
After each game, the loser must give the winner one state that is no longer absorbing. Determine whether
chip. Player 2 is more advanced than Player 1 and has a each matrix X below can be a steady state matrix for the
70% chance of winning each game. The tournament is Markov chain corresponding to P′. Explain.
over when one player obtains all four chips. What is the

[] []
6
probability that Player 1 will win the tournament? 11 0
5
47. Explain how you can determine the steady state matrix 11 0
X of an absorbing Markov chain by inspection. (a) X = (b) X = 5
0 11
6
0 11
48. CAPSTONE
(a) Explain how to find the nth state matrix of a 53. Proof Prove that the product of two 2 × 2 stochastic
Markov chain. matrices is stochastic.
(b) Explain how to find the steady state matrix of a 54. Proof Let P be a 2 × 2 stochastic matrix. Prove that
Markov chain. there exists a 2 × 1 state matrix X with nonnegative
entries such that PX = X.
(c) What is a regular Markov chain?
55. In Example 5, show that for the regular stochastic
(d) What is an absorbing Markov chain? matrix P, the sequence of successive powers
(e) How is an absorbing Markov chain different than a
regular Markov chain? P, P2, P3, . . .
approaches a stable matrix P, where the entries in each
49. Consider the Markov chain whose matrix of transition column of P are equal to the corresponding entries in the
probabilities P is given in Example 7(b). Show that the steady state matrix X. Repeat for several other regular
steady state matrix X depends on the initial state matrix stochastic matrices P and corresponding steady state
X0 by finding X for each X0. matrices X.
56. Proof Prove that when P is a regular stochastic

[ ] [ ]
0.25 0.25
matrix, the corresponding regular Markov chain
0.25 0.25
(a) X0 = (b) X0 =
0.25 0.40 PX0, P2X0, P3X0, . . .
0.25 0.10 approaches a unique steady state matrix X.

2.6 More applications of Matrix operations

Use matrix multiplication to encode and decode messages.
Use matrix algebra to analyze an economic system (Leontief
input-output model).
Find the least squares regression line for a set of data.

CryptograpHy
A cryptogram is a message written according to a secret code (the Greek word kryptos
means “hidden”). One method of using matrix multiplication to encode and decode
messages is introduced below.
To begin, assign a number to each letter in the alphabet (with 0 assigned to a blank
space), as shown.
0 = −− 9=I 18 = R
1=A 10 = J 19 = S
2=B 11 = K 20 = T
3=C 12 = L 21 = U
4=D 13 = M 22 = V
5=E 14 = N 23 = W
6=F 15 = O 24 = X
7=G 16 = P 25 = Y
8=H 17 = Q 26 = Z
Then convert the message to numbers and partition it into uncoded row matrices, each
having n entries, as demonstrated in Example 1.

forming uncoded row Matrices

Write the uncoded row matrices of size 1 × 3 for the message MEET ME MONDAY.
solution
Partitioning the message (including blank spaces, but ignoring punctuation) into groups
of three produces the uncoded row matrices shown below.
[13 5 5] [20 0 13] [5 0 13] [15 14 4] [1 25 0]
M E E T M E M O N D A Y
Note the use of a blank space to fill out the last uncoded row matrix.

Information security is of the utmost importance when

linEar conducting business online. If a malicious party should
algEBra receive confidential information such as passwords,
appliED personal identification numbers, credit card numbers,
Social Security numbers, bank account details, or sensitive
company information, then the effects can be damaging. To
protect the confidentiality and integrity of such information,
Internet security can include the use of data encryption, the
process of encoding information so that the only way to
decode it, apart from an “exhaustion attack,” is to use a key.
Data encryption technology uses algorithms based on the
material presented here, but on a much more sophisticated
level, to prevent malicious parties from discovering the key.
Andrea Danti/Shutterstock.com

To encode a message, choose an n × n invertible matrix A and multiply the

uncoded row matrices (on the right) by A to obtain coded row matrices. Example 2
demonstrates this process.

Encoding a Message

Use the invertible matrix

[ ]
1 −2 2
A = −1 1 3
1 −1 −4
to encode the message MEET ME MONDAY.
solution
Obtain the coded row matrices by multiplying each of the uncoded row matrices found
in Example 1 by the matrix A, as shown below.
Uncoded Encoding Coded Row
Row Matrix Matrix A Matrix

[ ]
1 −2 2
[13 5 5] −1 1 3 = [13 −26 21]
1 −1 −4

[ ]
1 −2 2
[20 0 13] −1 1 3 = [33 −53 −12]
1 −1 −4

[ ]
1 −2 2
[5 0 13] −1 1 3 = [18 −23 −42]
1 −1 −4

[ ]
1 −2 2
[15 14 4] −1 1 3 = [5 −20 56]
1 −1 −4

[ ]
1 −2 2
[1 25 0] −1 1 3 = [−24 23 77]
1 −1 −4
The sequence of coded row matrices is
[13 −26 21][33 −53 −12][18 −23 −42][5 −20 56][−24 23 77].
Finally, removing the matrix notation produces the cryptogram
13 −26 21 33 −53 −12 18 −23 −42 5 −20 56 −24 23 77.

For those who do not know the encoding matrix A, decoding the cryptogram
found in Example 2 is difficult. But for an authorized receiver who knows the
encoding matrix A, decoding is relatively simple. The receiver just needs to multiply
the coded row matrices by A−1 to retrieve the uncoded row matrices. In other words, if
X = [x1 x2 . . . xn ]
is an uncoded 1 × n matrix, then Y = XA is the corresponding encoded matrix. The
receiver of the encoded matrix can decode Y by multiplying on the right by A−1 to
obtain
YA−1 = (XA)A−1 = X.
Example 3 demonstrates this procedure.

Decoding a Message

Use the inverse of the matrix

[ ]
1 −2 2
A = −1 1 3
1 −1 −4
to decode the cryptogram
13 −26 21 33 −53 −12 18 −23 −42 5 −20 56 −24 23 77.

solution
Begin by using Gauss-Jordan elimination to find A−1.
[A I] [I A−1]

[ ] [ ]
1 −2 2 1 0 0 1 0 0 −1 −10 −8
−1 1 3 0 1 0 0 1 0 −1 −6 −5
1 −1 −4 0 0 1 0 0 1 0 −1 −1
Now, to decode the message, partition the message into groups of three to form the
coded row matrices
[13 −26 21][33 −53 −12][18 −23 −42][5 −20 56][−24 23 77].
To obtain the decoded row matrices, multiply each coded row matrix by A−1
(on the right).
Coded Row Decoding Decoded
Matrix Matrix A−1 Row Matrix

[ ]
−1 −10 −8
[13 −26 21] −1 −6 −5 = [13 5 5]
0 −1 −1

[ ]
−1 −10 −8
[33 −53 −12] −1 −6 −5 = [20 0 13]
0 −1 −1

[ ]
−1 −10 −8
[18 −23 −42] −1 −6 −5 = [5 0 13]
0 −1 −1

[ ]
−1 −10 −8
[5 −20 56] −1 −6 −5 = [15 14 4]
0 −1 −1

[ ]
−1 −10 −8
[−24 23 77] −1 −6 −5 = [1 25 0]
0 −1 −1
The sequence of decoded row matrices is
[13 5 5][20 0 13][5 0 13][15 14 4][1 25 0]
and the message is
13 5 5 20 0 13 5 0 13 15 14 4 1 25 0.
M E E T M E M O N D A Y

Leontief Input-Output Models

In 1936, American economist Wassily W. Leontief (1906–1999) published a model
concerning the input and output of an economic system. In 1973, Leontief received a
Nobel prize for his work in economics. A brief discussion of Leontief’s model follows.
Consider an economic system that has n different industries I1, I2, . . . , In , each
having input needs (raw materials, utilities, etc.) and an output (finished product).
In producing each unit of output, an industry may use the outputs of other industries,
including itself. For example, an electric utility uses outputs from other industries, such
as coal and water, and also uses its own electricity.
Let dij be the amount of output the jth industry needs from the ith industry to produce
one unit of output per year. The matrix of these coefficients is the input-output matrix.
User (Output)
I1 I2 . . . In

[ ]
d11 d12 . . . d1n I1
d21 d22 . . . d2n I2
D= Supplier (Input)
⋮ ⋮ ⋮ ⋮
dn1 dn2 . . . dnn In
To understand how to use this matrix, consider d12 = 0.4. This means that
for Industry 2 to produce one unit of its product, it must use 0.4 unit of Industry 1’s
product. If d33 = 0.2, then Industry 3 needs 0.2 unit of its own product to produce one
unit. For this model to work, the values of dij must satisfy 0 ≤ dij ≤ 1 and the sum of
the entries in any column must be less than or equal to 1.

Forming an Input-Output Matrix

Consider a simple economic system consisting of three industries: electricity, water,

and coal. Production, or output, of one unit of electricity requires 0.5 unit of itself,
0.25 unit of water, and 0.25 unit of coal. Production of one unit of water requires
0.1 unit of electricity, 0.6 unit of itself, and 0 units of coal. Production of one unit of
coal requires 0.2 unit of electricity, 0.15 unit of water, and 0.5 unit of itself. Find the
input-output matrix for this system.
solution
The column entries show the amounts each industry requires from the others, and from
itself, to produce one unit of output.
User (Output)
E W C

[ ]
0.5 0.1 0.2 E
0.25 0.6 0.15 W Supplier (Input)
0.25 0 0.5 C
The row entries show the amounts each industry supplies to the others, and to itself,
for that industry to produce one unit of output. For instance, the electricity industry
supplies 0.5 unit to itself, 0.1 unit to water, and 0.2 unit to coal.

To develop the Leontief input-output model further, let the total output of the
ith industry be denoted by xi . If the economic system is closed (that is, the economic
system sells its products only to industries within the system, as in the example above),
then the total output of the ith industry is
xi = di1x1 + di2 x2 + . . . + din xn. Closed system

On the other hand, if the industries within the system sell products to nonproducing
groups (such as governments or charitable organizations) outside the system, then the
system is open and the total output of the ith industry is
xi = di1x1 + di2 x2 + . . . + din xn + ei Open system

where ei represents the external demand for the ith industry’s product. The system of
n linear equations below represents the collection of total outputs for an open system.
x1 = d11x1 + d12 x2 + . . . + d1n xn + e1
x2 = d21x1 + d22 x2 + . . . + d2n xn + e2
⋮
xn = dn1x1 + dn2 x2 + . . . + dnn xn + en
The matrix form of this system is X = DX + E, where X is the output matrix and E
is the external demand matrix.

solving for the output Matrix

of an open system
See LarsonLinearAlgebra.com for an interactive version of this type of example.

An economic system composed of three industries has the input-output matrix shown
below.
User (Output)
A B C

[ ]
0.1 0.43 0 A
D = 0.15 0 0.37 B Supplier (Input)
0.23 0.03 0.02 C
Find the output matrix X when the external demands are

[ ]
20,000 A
E = 30,000 . B
25,000 C

solution
Letting I be the identity matrix, write the equation X = DX + E as IX − DX = E,
which means that (I − D)X = E. Using the matrix D above produces

[ ]
0.9 − 0.43 0
rEMarK I − D = − 0.15 1 − 0.37 .
The economic systems − 0.23 − 0.03 0.98
described in Examples 4 and 5
are, of course, simple ones. Using Gauss-Jordan elimination,
In the real world, an economic

[ ]
1.25 0.55 0.21
system would include many
industries or industrial groups. (I − D) −1
≈ 0.30 1.14 0.43 .
A detailed analysis using 0.30 0.16 1.08
the Leontief input-output
So, the output matrix is
model could easily require

[ ][ ] [ ]
an input-output matrix greater 1.25 0.55 0.21 20,000 46,750 A
than 100 × 100 in size. Clearly, X = (I − D)−1E ≈ 0.30 1.14 0.43 30,000 = 50,950 B
this type of analysis would
0.30 0.16 1.08 25,000 37,800 C
require the aid of a computer.
To produce the given external demands, the outputs of the three industries must be
approximately 46,750 units for industry A, 50,950 units for industry B, and 37,800 units
for industry C.

Least Squares Regression Analysis

You will now look at a procedure used in statistics to develop linear models. The next
example demonstrates a visual method for approximating a line of best fit for a set of
data points.

A Visual Straight-Line Approximation

Determine a line that appears to best fit the points (1, 1), (2, 2), (3, 4), (4, 4), and (5, 6).
solution
Plot the points, as shown in Figure 2.5. It appears that a good choice would be the line
whose slope is 1 and whose y-intercept is 0.5. The equation of this line is
y = 0.5 + x.

An examination of the line in Figure 2.5 reveals that you can improve the fit
by rotating the line counterclockwise slightly, as shown in Figure 2.6. It seems clear
that this line, whose equation is y = 1.2x, fits the points better than the original line.

y y

6 (5, 6) 6 y = 1.2x (5, 6)

5 5

4 (3, 4) (4, 4) 4 (3, 4) (4, 4)

3 3
y = 0.5 + x y = 0.5 + x
2 (2, 2) 2 (2, 2)
1 (1, 1) 1 (1, 1)
y
x x
6 Model 1 (5, 6) 1 2 3 4 5 6 1 2 3 4 5 6

5 Figure 2.5 Figure 2.6

4 (3, 4) (4, 4) One way of measuring how well a function y = f (x) fits a set of points
3 (x1, y1), (x2, y2 ), . . . , (xn, yn )
y = 0.5 + x
2 (2, 2) is to compute the differences between the values from the function f (xi ) and the
1 (1, 1) actual values yi . These values are shown in Figure 2.7. By squaring the differences and
summing the results, you obtain a measure of error called the sum of squared error.
x The table shows the sums of squared errors for the two linear models.
1 2 3 4 5 6

y
Model 1: f (x) = 0.5 + x Model 2: f (x) = 1.2x
6 Model 2 (5, 6)
xi yi f (xi ) [ yi − f (xi )]2 xi yi f (xi ) [ yi − f (xi )]2
5

4 (3, 4) (4, 4)
1 1 1.5 (−0.5)2 1 1 1.2 (−0.2)2

3
2 2 2.5 (−0.5)2 2 2 2.4 (−0.4)2
y = 1.2x
2 (2, 2) 3 4 3.5 (+0.5)2 3 4 3.6 (+0.4)2

1 (1, 1) 4 4 4.5 (−0.5)2 4 4 4.8 (−0.8)2

x 5 6 5.5 (+0.5)2 5 6 6.0 (0.0)2
1 2 3 4 5 6
Sum 1.25 Sum 1.00
Figure 2.7

The sums of squared errors confirm that the second model fits the points better
than the first model.
Of all possible linear models for a given set of points, the model that has the best
fit is the one that minimizes the sum of squared error. This model is the least squares
regression line, and the procedure for finding it is the method of least squares.

Definition of Least Squares Regression Line

For a set of points
(x1, y1), (x2, y2), . . . , (xn yn )
the least squares regression line is the linear function
f (x) = a0 + a1x
that minimizes the sum of squared error
[ y1 − f (x1)]2 + [ y2 − f (x2 )]2 + . . . + [ yn − f (xn )]2.

To find the least squares regression line for a set of points, begin by forming the
system of linear equations
y1 = f (x1) + [ y1 − f (x1)]
y2 = f (x2 ) + [ y2 − f (x2 )]
⋮
yn = f (xn ) + [ yn − f (xn )]
where the right-hand term
[ yi − f (xi )]
of each equation is the error in the approximation of yi by f (xi ). Then write this
error as
ei = yi − f (xi )
and write the system of equations in the form
y1 = (a0 + a1x1) + e1
y2 = (a0 + a1x2 ) + e2
⋮
yn = (a0 + a1xn ) + en.
Now, if you define Y, X, A, and E as

[] [ ] []
y1 1 x1 e1
Y=
y2
⋮
, X =
⋮
1 x2
⋮
, A =
a0
a1
e
, E = 2
⋮ [ ]
yn 1 xn en
then the n linear equations may be replaced by the matrix equation
Y = XA + E.
Note that the matrix X has a column of 1’s (corresponding to a0) and a column
containing the xi’s. This matrix equation can be used to determine the coefficients of
the least squares regression line, as shown on the next page.

Matrix Form for Linear Regression

For the regression model Y = XA + E, the coefficients of the least squares
REMARK regression line are given by the matrix equation
You will learn more about this A = (X T X)−1X TY
procedure in Section 5.4.
and the sum of squared error is E TE.

Example 7 demonstrates the use of this procedure to find the least squares
regression line for the set of points from Example 6.

Finding the Least Squares Regression Line

Find the least squares regression line for the points (1, 1), (2, 2), (3, 4), (4, 4),
and (5, 6).
solution
The matrices X and Y are

[ ] []
1 1 1
1 2 2
X= 1 3 and Y = 4 .
1 4 4
1 5 6
This means that

[ ]
1 1
1 2
X TX = [11 1
2
1
3
1
4
1
5 ] 1 3 =
5
15 [ 15
55 ]
1 4
1 5
and

[]
1
2
X TY = [11 1
2
1
3
1
4
1
5 ] 4 =
17
63
. [ ]
y 4
6
6 (5, 6)
Now, using (X TX )−1 to find the coefficient matrix A, you have
5
A = (X TX)−1X TY
4 (3, 4) (4, 4)
−15
3 y = − 0.2 + 1.2x
1
[−15
= 50
55
5 ][17
63]
(2, 2) −0.2
=[
1.2]
2
.
1 (1, 1)
x So, the least squares regression line is
1 2 3 4 5 6
Least Squares Regression Line y = −0.2 + 1.2x
Figure 2.8 as shown in Figure 2.8. The sum of squared error for this line is 0.8 (verify this), which
means that this line fits the data better than either of the two experimental linear
models determined earlier.

2.6 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

Encoding a Message In Exercises 1 and 2, write the 9. Decoding a Message Use a software program or a
uncoded row matrices for the message. Then encode the graphing utility to decode the cryptogram.
message using the matrix A.

[ ]
1 0 2
1. Message: SELL CONSOLIDATED A = 2 −1 1
Row Matrix Size: 1×3 0 1 2

[ ]
1 −1 0 38 −14 29 56 −15 62 17 3 38 18 20 76 18 −5
Encoding Matrix: A= 1 0 −1 21 29 −7 32 32 9 77 36 −8 48 33 −5 51 41
−6 2 3 3 79 12 1 26 58 −22 49 63 −19 69 28 8 67 31
−11 27 41 −18 28
2. Message: HELP IS COMING
10. Decoding a Message A code breaker intercepted
Row Matrix Size: 1×4
the encoded message below.

[ ]
−2 3 −1 −1 45 −35 38 −30 18 −18 35 −30 81 −60 42
−1 1 1 1 −28 75 −55 2 −2 22 −21 15 −10
Encoding Matrix: A=
−1 −1 1 2
Let the inverse of the encoding matrix be
3 1 −2 −4
Decoding a Message In Exercises 3–6, use A−1 to
A−1 =
w
y
x
z [
. ]
decode the cryptogram. (a) You know that [45 −35] A−1 = [10 15] and
[38 −30] A−1 = [8 14]. Write and solve two
3. A =
1
3 [ 2
5
, ] systems of equations to find w, x, y, and z.
11 21 64 112 25 50 29 53 23 46 40 75 55 92 (b) Decode the message.
11. Industrial System A system composed of two
4. A = [23
3
4
, ] industries, coal and steel, has the input requirements
below.
85 120 6 8 10 15 84 117 42 56 90 125 60 80
30 45 19 26 (a) To produce $1.00 worth of output, the coal industry
requires $0.10 of its one product and $0.80 of steel.

[ ]
1 2 2
(b) To produce $1.00 worth of output, the steel industry
5. A = 3 7 9 ,
requires $0.10 of its own product and $0.20 of coal.
−1 −4 −7
Find D, the input-output matrix for this system.
13 19 10 −1 −33 −77 3 −2 −14 4 1 −9 −5 Then solve for the output matrix X in the equation
−25 −47 4 1 −9 X = DX + E, where E is the external demand matrix

[ ]
3 −4 2
[ ]
10,000
E= .
6. A = 0 2 1 , 20,000
4 −5 3 12. Industrial System An industrial system has two
112 −140 83 19 −25 13 72 −76 61 95 −118 industries with the input requirements below.
71 20 21 38 35 −23 36 42 −48 32 (a) To produce $1.00 worth of output, Industry A
requires $0.30 of its own product and $0.40 of
7. Decoding a Message The cryptogram below was Industry B’s product.
encoded with a 2 × 2 matrix. The last word of the
(b) To produce $1.00 worth of output, Industry B
message is __RON. What is the message?
requires $0.20 of its own product and $0.40 of
8 21 −15 −10 −13 −13 5 10 5 25 5 19 −1 6 Industry A’s product.
20 40 −18 −18 1 16
Find D, the input-output matrix for this system.
8. Decoding a Message The cryptogram below was Then solve for the output matrix X in the equation
encoded with a 2 × 2 matrix. The last word of the X = DX + E, where E is the external demand matrix
message is __SUE. What is the message?
5 2 25 11 −2 −7 −15 −15 32 14 −8 −13 38 E=
10,000
20,000 [. ]
19 −19 −19 37 16

13. Solving for the Output Matrix A small community 27. Demand A hardware retailer wants to know the
includes a farmer, a baker, and a grocer and has the demand for a rechargeable power drill as a function
input-output matrix D and external demand matrix E of price. The ordered pairs (25, 82), (30, 75), (35, 67),
below. and (40, 55) represent the price x (in dollars) and the
Farmer Baker Grocer corresponding monthly sales y.
(a) Find the least squares regression line for the data.

[ ] [ ]
0.4 0.5 0.5 Farmer 1000
D = 0.3 0.0 0.3 Baker and E = 1000 (b) Estimate the demand when the price is $32.95.
0.2 0.2 0.0 Grocer 1000 28. Wind Energy Consumption The table shows the
Solve for the output matrix X in the equation wind energy consumptions y (in quadrillions of Btus,
X = DX + E. or British thermal units) in the United States from
2009 through 2013. Find the least squares regression
14. Solving for the Output Matrix An industrial system
line for the data. Let t represent the year, with t = 9
with three industries has the input-output matrix D and
corresponding to 2009. Use the linear regression
external demand matrix E below.
capabilities of a graphing utility to check your work.

[ ] [ ]
0.2 0.4 0.4 5000 (Source: U.S. Energy Information Administration)
D= 0.4 0.2 0.2 and E = 2000
0.0 0.2 0.2 8000 Year 2009 2010 2011 2012 2013

Solve for the output matrix X in the equation Consumption, y 0.72 0.92 1.17 1.34 1.60
X = DX + E.
29. Wildlife A wildlife management team studied the
Least Squares Regression Analysis In Exercises reproduction rates of deer in three tracts of a wildlife
15–18, (a) sketch the line that appears to be the best fit preserve. The team recorded the number of females x in
for the given points, (b) find the least squares regression each tract and the percent of females y in each tract that
line, and (c) determine the sum of squared error. had offspring the following year. The table shows the
y y
15. 16. results.
4 4
(2, 3) 3 Number, x 100 120 140
3 (3, 2)
2
(−1, 1) Percent, y 75 68 55
2 (1, 1)
x
(−2, 0)
1
(0, 1) (a) Find the least squares regression line for the data.
(− 3, 0) − 1 1 2 3
x
−2
(b) Use a graphing utility to graph the model and the
−2 −1 1 2
data in the same viewing window.
17. y
18. y (c) Use the model to create a table of estimated values for
(0, 4) y. Compare the estimated values with the actual data.
4 4 (5, 2)
3 (4, 2) (6, 2) (d) Use the model to estimate the percent of females
3 (3, 1)
(1, 3) 2 that had offspring when there were 170 females.
2 (1, 0)
1 (4, 1) (e) Use the model to estimate the number of females
x
1
5 6
when 40% of the females had offspring.
(1, 1) (2, 0) −1
x (2, 0) (3, 0)
−1 1 2 3 −2
30. CAPSTONE
Finding the Least Squares Regression Line In (a) Explain how to use matrix multiplication to encode
Exercises 19–26, find the least squares regression line. and decode messages.
19. (0, 0), (1, 1), (2, 4) (b) Explain how to use a Leontief input-output model
20. (1, 0), (3, 3), (5, 6) to analyze an economic system.
21. (−2, 0), (−1, 1), (0, 1), (1, 2) (c) Explain how to use matrices to find the least
22. (−4, −1), (−2, 0), (2, 4), (4, 5) squares regression line for a set of data.
23. (−5, 1), (1, 3), (2, 3), (2, 5)
24. (−3, 4), (−1, 2), (1, 1), (3, 0) 31. Use your school’s library, the Internet, or some other
25. (−5, 10), (−1, 8), (3, 6), (7, 4), (5, 5) reference source to derive the matrix form for linear
26. (0, 6), (4, 3), (5, 0), (8, −4), (10, −5) regression given at the top of page 101.

2 Review Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

Operations with Matrices In Exercises 1–6, perform Using the Inverse of a Matrix In Exercises 19–26,
the matrix operations. use an inverse matrix to solve each system of linear
−6 equations or matrix equation.
1.
2
0[ 1
5 −4
0
−3
5
0 ] [ 3
−2 5 ] 19. 5x1 + 4x2 = 2 20. 3x1 + 2x2 = 1
−x1 + x2 = −22 x1 + 4x2 = −3

[ ] [ ]
1 2 7 1
2. −2 5 −4 + 8 1 2 21. −x1 + x2 + 2x3 = 1
6 0 1 4 2x1 + 3x2 + x3 = −2
5x1 + 4x2 + 2x3 = 4

[ ][
1 2
−2
3. 5 ]
−4
6
4 0
8
0
22. x1 + x2 + 2x3 = 0
x1 − x2 + x3 = −1
6 0
2x1 + x2 + x3 = 2
−2
4. [12 5
] [64 8
]
−4 0 0 23. [−15 4
1 ] [xy] = [−15
−6]

[ ][ ]
1 3 2 4 −3 2
−1
5. 0
0
2
0
−4
3
0
0
3
0
−1
2
24. [23 4 ] [xy] = [−25]

[ ][ ] [ ]
−2 0 1 −2 x1 −1
6. [26 1
0 ] [−34 2
1
+] [
0
4
4 ] 25. −1 3 1 x2 = 0
2 −2 4 x3 2
Solving a System of Linear Equations In Exercises

[ ][ ] [ ]
0 1 2 x 0
7–10, write the system of linear equations in the form
Ax = b. Then use Gaussian elimination to solve this 26. 3 2 1 y = −1
matrix equation for x. 4 −3 −4 z −7
7. 2x1 + x2 = −8 8. 2x1 − x2 = 5 Solving a Matrix Equation In Exercises 27 and 28,
x1 + 4x2 = −4 3x1 + 2x2 = −4 find A.
−1
9. −3x1 − x2 + x3 = 0
2x1 + 4x2 − 5x3 = −3
27. (3A)−1 = [42 3 ] 28. (2A)−1 = [20 4
1 ]
x1 − 2x2 + 3x3 = 1
Nonsingular Matrix In Exercises 29 and 30, find x
10. 2x1 + 3x2 + x3 = 10 such that the matrix A is nonsingular.
2x1 − 3x2 − 3x3 = 22
4x1 − 2x2 + 3x3 = −2 29. A =
3
x −1
1
[ 30. A = ]2
1
x
4 [ ]
Finding and Multiplying with a Transpose Finding the Inverse of an Elementary Matrix
In Exercises 11–14, find AT, ATA, and AAT. In Exercises 31 and 32, find the inverse of the elementary
2 −3 3 −1 matrix.
11. A =
1
[ 12. A = ] [ ]
[ ] [ ]
0 1 2 2 0 1 0 4 1 0 0
31. 0 1 0 32. 0 6 0

[ ]
1
13. A = 3 14. A = [1 −2 −3] 0 0 1 0 0 1
−1 Finding a Sequence of Elementary Matrices
In Exercises 33–36, find a sequence of elementary
Finding the Inverse of a Matrix In Exercises 15–18,
matrices whose product is the given nonsingular matrix.
find the inverse of the matrix (if it exists).
−3 13
15.
3 −1
[ ] 16.
4 −1
[ ]
33.
2
0 [3
1 ]
34.
1 −4 [ ]
2 −1 −8 2

[ ] [ ]
1 0 1 3 0 6

[ ] [ ]
2 3 1 1 1 1
35. 0 1 −2 36. 0 2 0
17. 2 −3 −3 18. 0 1 1
0 0 4 1 0 3
4 0 3 0 0 1

37. Find two 2 × 2 matrices A such that A2 = I. 50. Manufacturing A corporation has four factories,
38. Find two 2 × 2 matrices A such that A2 = O. each of which manufactures sport utility vehicles and
pickup trucks. In the matrix
39. Find three 2 × 2 idempotent matrices. (Recall that a
square matrix A is idempotent when A2 = A.) A= [
100 90 70 30
40 20 60 60 ]
40. Find 2 × 2 matrices A and B such that AB = O but
BA ≠ O. aij represents the number of vehicles of type i produced
at factory j in one day. Find the production levels when
41. Consider the matrices below.
production increases by 10%.

[] [ ] [ ] [ ]
1 −1 3 3 51. Gasoline Sales Matrix A shows the numbers of
2 0 4 2 gallons of 87-octane, 89-octane, and 93-octane gasoline
X= , Y= , Z= , W=
0 3 −1 −4 sold at a convenience store over a weekend.
1 2 2 −1 Octane
(a) Find scalars a, b, and c such that W = aX + bY + cZ.
(b) Show that there do not exist scalars a and b such 87 89 93

[ ]
that Z = aX + bY. 580 840 320 Friday
(c) Show that if aX + bY + cZ = O, then a = b = c = 0. A = 560 420 160 Saturday
42. Proof Let A, B, and A + B be nonsingular matrices. 860 1020 540 Sunday
Prove that A−1 + B−1 is nonsingular by showing that Matrix B gives the selling prices (in dollars per gallon)
and the profits (in dollars per gallon) for the three
(A−1 + B−1 −1 ) = A(A + B) −1B.
grades of gasoline.
Finding an LU-Factorization of a Matrix In Exercises Selling Price Profit

[ ]
43–46, find an LU-factorization of the matrix. b11 0.05 87
B = b21 0.08 89 Octane
43. [26 5
14 ] 44. [−3
12
1
0 ] b31 0.10 93
(a) Find AB and interpret the result.

[ ] [ ]
4 1 0 1 1 1
45. 0 3 −7 46. 1 2 2 (b) Find the convenience store’s profit from gasoline
−16 11 1 1 2 3 sales for the weekend.
52. Final Grades Two midterms and a final exam
Solving a Linear System Using LU-Factorization determine the final grade in a course at a liberal arts
In Exercises 47 and 48, use an LU-factorization of the college. The matrices below show the grades for six
coefficient matrix to solve the linear system. students and two possible grading systems.
47. x + z=3 Midterm Midterm Final
2x + y + 2z = 7 1 2 Exam
3x + 2y + 6z = 8

[ ]
78 82 80 Student 1
48. 2x1 + x2 + x3 − x4 = 7 84 88 85 Student 2
3x2 + x3 − x4 = −3 92 93 90 Student 3
−2x3 = 2 A=
88 86 90 Student 4
2x1 + x2 + x3 − 2x4 = 8 74 78 80 Student 5
49. Manufacturing A company manufactures tables and 96 95 98 Student 6
chairs at two locations. Matrix C gives the costs of
Grading Grading
manufacturing at each location.
System 1 System 2
Location 1 Location 2

[ ]
0.25 0.20 Midterm 1
C=
627
135 [ 681 Tables
150 Chairs ] B = 0.25
0.50
0.20 Midterm 2
0.60 Final Exam
2
(a) Labor accounts for of the cost. Determine the 3 (a) Describe the grading systems in matrix B.
matrix L that gives the labor costs at each location.
(b) Compute the numerical grades for the six students
(b)
Find the matrix M that gives material costs at
(to the nearest whole number) using the two grading
each location. (Assume there are only labor and
systems.
material costs.)
(c) How many students received an “A” in each
grading system? (Assume 90 or greater is an “A.”)

Polynomial Function In Exercises 53 and 54, find f (A) Regular and Steady State Matrix In Exercises 65–68,
using the definition below. determine whether the stochastic matrix P is regular.
If f (x) = a0 + a1x + a2 x2 + . . . + an xn is a polynomial Then find the steady state matrix X of the Markov chain
function, then for a square matrix A, with matrix of transition probabilities P.

[ ]
4
f (A) = a0 I + a1 A + a2 A2 + . . . + an An. 1
65. P = [ 0.8
0.2
0.5
0.5 ] 66. P =
0
7
3
5
[ 4
]
7
53. f (x) = 6 − 7x + x2, A=

[ ]
1 1
1 2 0

[ ]
3 6 0 0 0.2
1
67. P = 0 0 68. P = 0.5 0.9 0
54. f (x) = 2 − 3x + x3, A= [−12 1
0 ] 6
1
2
5
6 1 0.5 0.1 0.8

Stochastic Matrices In Exercises 55–58, determine 69. S ales Promotion As a promotional feature, a store
whether the matrix is stochastic. conducts a weekly raffle. During any week, 40% of the

[ ]
12 2
customers who turn in one or more tickets do not bother
55.
25
13
25
25
23
25
56. [0.30 0.7
1 ] to turn in tickets the following week. On the other hand,
30% of the customers who do not turn in tickets will

[ ] [ ]
1 0 0 0.3 0.4 0.1 turn in one or more tickets the following week. Find and
57. 0 0.5 0.1 58. 0.2 0.4 0.5 interpret the steady matrix for this situation.
0 0.1 0.5 0.5 0.2 0.4 70. Classified Documents A courtroom has 2000
documents, of which 1250 are classified. Each week,
Finding State Matrices In Exercises 59–62, use the 10% of the classified documents become declassified
matrix of transition probabilities P and initial state and 20% are shredded. Also, 20% of the unclassified
matrix X0 to find the state matrices X1, X2, and X3. documents become classified and 5% are shredded.
Find and interpret the steady state matrix for this
[ ] []
1 1 2
2 4 3
59. P = 1 3 , X0 = 1 situation.
2 4 3
Absorbing Markov Chains In Exercises 71 and 72,
60. P = [ 0.23 0.45
0.77 0.55
, ] X0 = [ ]0.65
0.35
determine whether the Markov chain with matrix of
transition probabilities P is absorbing. Explain.

[ ] [ ]
0.50 0.25 0 0.5

[ ] [ ]
0 0.4 0.1 1 0 0.38
61. P = 0.25 0.70 0.15 , X0 = 0.5
71. P = 0.7 0.3 0.4 72. P = 0 0.30 0
0.25 0.05 0.85 0
0.3 0.3 0.5 0 0.70 0.62

[ ] []
1 1 2 2
3 3 3 9
1 1 4 True or False? In Exercises 73–76, determine whether
62. P = 3 0 3 , X0 = 9 each statement is true or false. If a statement is true, give
1 2 1
3 3 0 3
a reason or cite an appropriate statement from the text.
If a statement is false, provide an example that shows the
63. Caribbean Cruise Three hundred people go on a statement is not true in all cases or cite an appropriate
Caribbean cruise. When the ship stops at a port, each statement from the text.
person has a choice of going on shore or not. Seventy 73. (a) Addition of matrices is not commutative.
percent of the people who go on shore one day will not
go on shore the next day. Sixty percent of the people (b) The transpose of the sum of matrices is equal to the
who do not go on shore one day will go on shore the sum of the transposes of the matrices.
next day. Today, 200 people go on shore. How many 74. (a) If an n × n matrix A is not symmetric, then ATA is
people will go on shore (a) tomorrow and (b) the day not symmetric.
after tomorrow? (b) If A and B are nonsingular n × n matrices, then
64. Population Migration A country has three regions. A + B is a nonsingular matrix.
Each year, 10% of the residents of Region 1 move 75. (a) A stochastic matrix can have negative entries.
to Region 2 and 5% move to Region 3, 15% of the (b) A Markov chain that is not regular can have a
residents of Region 2 move to Region 1 and 5% move unique steady state matrix.
to Region 3, and 10% of the residents of Region 3 move
to Region 1 and 10% move to Region 2. This year, each 76. (a) A regular stochastic matrix can have entries of 0.
region has a population of 100,000. Find the populations (b) The steady state matrix of an absorbing Markov
of each region (a) in 1 year and (b) in 3 years. chain always depends on the initial state matrix.

Encoding a Message In Exercises 77 and 78, write the 84. Solving for the Output Matrix An industrial system
uncoded row matrices for the message. Then encode the with three industries has the input-output matrix D and
message using the matrix A. external demand matrix E below.
77. Message: ONE IF BY LAND

[ ] [ ]
0.1 0.3 0.2 3000
Row Matrix Size: 1×2 D = 0.0 0.2 0.3 and E = 3500
0.4 0.1 0.1 8500
Encoding Matrix: A=
5
2
2
1 [ ]
Solve for the output matrix X in the equation
X = DX + E.
78. Message: BEAM ME UP SCOTTY
Row Matrix Size: 1×3 Finding the Least Squares Regression Line In
Exercises 85–88, find the least squares regression line.

[ ]
2 1 4
85. (1, 5), (2, 4), (3, 2)
Encoding Matrix: A= 3 1 3
−2 −1 −3 86. (2, 1), (3, 3), (4, 2), (5, 4), (6, 4)
87. (1, 1), (1, 3), (1, 2), (1, 4), (2, 5)
Decoding a Message In Exercises 79–82, use A−1 to
88. (−2, 4), (−1, 2), (0, 1), (1, −2), (2,−3)
decode the cryptogram.
−2 89. Cellular Phone Subscribers The table shows the
79. A =
3
, [−43 ] numbers of cellular phone subscribers y (in millions)
in the United States from 2008 through 2013. (Source:
− 45 34 36 −24 −43 37 −23 22 −37 29 57 −38

CTIA–The Wireless Association)
−39 31
Year 2008 2009 2010 2011 2012 2013
80. A =
−3
4
, [−11 ] Number, y 270 286 296 316 326 336
11 52 −8 −9 −13 −39 5 20 12 56 5 20 −2 7
9 41 25 100 (a) Find the least squares regression line for the data.
Let x represent the year, with x = 8 corresponding

[ ]
1 −2 2
to 2008.
81. A = −1 1 3
1 −1 −4 (b) Use the linear regression capabilities of a graphing
utility to find a linear model for the data. How does
− 2 2 5 39 −53 −72 −6 −9 93 4 −12 27 31
this model compare with the model obtained in
−49 −16 19 −24 −46 −8 −7 99 part (a)?
(c) Use the linear model to create a table of estimated

[ ]
2 0 1
82. A = 2 −1 0 values for y. Compare the estimated values with the
1 2 −4 actual data.
90. Major League Baseball Salaries The table shows
66 27 −31 37 5 −9 61 46 −73 46 −14 9 94 the average salaries y (in millions of dollars) of Major
21 −49 32 −4 12 66 31 −53 47 33 −67 32 19 League Baseball players on opening day of baseball
−56 43 −9 −20 68 23 −34 season from 2008 through 2013. (Source: Major League
83. I ndustrial System An industrial system has two Baseball)
industries with the input requirements below.
Year 2008 2009 2010 2011 2012 2013
(a) To produce $1.00 worth of output, Industry A
requires $0.20 of its own product and $0.30 of Salary, y 2.93 3.00 3.01 3.10 3.21 3.39
Industry B’s product.
(b) To produce $1.00 worth of output, Industry B (a) Find the least squares regression line for the data.
requires $0.10 of its own product and $0.50 of Let x represent the year, with x = 8 corresponding
Industry A’s product. to 2008.
Find D, the input-output matrix for this system. (b) Use the linear regression capabilities of a graphing
Then solve for the output matrix X in the equation utility to find a linear model for the data. How does
X = DX + E, where E is the external demand matrix this model compare with the model obtained in
part (a)?
E=
40,000
80,000
. [ ] (c) Use the linear model to create a table of estimated
values for y. Compare the estimated values with the
actual data.

2 Projects

1 Exploring Matrix Multiplication

The table shows the first two test scores for Anna, Bruce, Chris, and David. Use the
Test 1 Test 2 table to create a matrix M to represent the data. Input M into a software program or
Anna 84 96 a graphing utility and use it to answer the questions below.
1. Which test was more difficult? Which was easier? Explain.
Bruce 56 72
2. How would you rank the performances of the four students?
Chris 78 83
David 82 91
3. Describe the meanings of the matrix products M
1
0
and M
0
1
. [] []
4. Describe the meanings of the matrix products [1 0 0 0]M and [0 0 1 0]M.

5. Describe the meanings of the matrix products M

1
1
and 12M
1
1
. [] []
6. Describe the meanings of the matrix products [1 1 1 1]M and 14 [1 1 1 1]M.

7. Describe the meaning of the matrix product [1 1 1 1]M

1
1
. []
8. Use matrix multiplication to find the combined overall average score on
both tests.
9. How could you use matrix multiplication to scale the scores on test 1 by a
factor of 1.1?

2 Nilpotent Matrices
Let A be a nonzero square matrix. Is it possible that a positive integer k exists such
that Ak = O? For example, find A3 for the matrix

[ ]
0 1 2
A= 0 0 1 .
0 0 0

A square matrix A is nilpotent of index k when A ≠ O, A2 ≠ O, . . . , Ak−1 ≠ O,

but Ak = O. In this project you will explore nilpotent matrices.
1. The matrix in the example above is nilpotent. What is its index?
2. Use a software program or a graphing utility to determine which matrices below
are nilpotent and find their indices.

(a)
0
0 [1
0
(b) ] 0
1
1
0
(c)
0
1 [ 0
0 ] [ ]

[ ] [ ]
0 0 1 0 0 0
(d) [ 1
1
0
0 ] (e) 0 0 0 (f) 1 0 0
0 0 0 1 1 0
3. Find 3 × 3 nilpotent matrices of indices 2 and 3.
4. Find 4 × 4 nilpotent matrices of indices 2, 3, and 4.
5. Find a nilpotent matrix of index 5.
6. Are nilpotent matrices invertible? Prove your answer.
7. When A is nilpotent, what can you say about AT? Prove your answer.
8. Show that if A is nilpotent, then I − A is invertible.

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Matric S

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2 Matrices

2.1 Operations with Matrices

Data Encryption (p. 94)

Computational Fluid Dynamics (p. 79)

Beam Deflection (p. 64)

Information Retrieval (p. 58)

Flight Crew Scheduling (p. 47)

2.1 operations with Matrices

definition of Equality of Matrices

Matrix Addition, Subtraction,

Definition of Matrix Addition

Scalar Multiplication and Matrix Subtraction

Definition of Matrix Multiplication

= ai1b1j + ai2b2j + ai3b3j + . . . + ainbnj.

finding the product of two Matrices

Find the product AB, where

Photo Researchers/Getty Images

ai1b1j + ai2b2j + ai3b3j + . . . + ainbnj = cij

1. Find A + B and B + A. Is matrix addition commutative?

2. Find AB and BA. Is matrix multiplication commutative?

See LarsonLinearAlgebra.com for an interactive version of this type of example.

Systems of Linear Equations

Solving a System of Linear Equations

Solve the matrix equation Ax = 0, where

Linear Combinations of Column Vectors

Solving a System of linear Equations

The linear system

Many real-life applications of linear systems involve

2.1 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

2.2 Properties of Matrix Operations

Theorem 2.1 Properties of Matrix Addition

Addition of More than Two Matrices

solving a Matrix equation

Solve for X in the equation 3X + A = B, where

X = 13 ([−32 41] − [10 −2

Properties of Matrix Multiplication

Matrix Multiplication Is Associative

Noncommutativity of Matrix Multiplication

Show that AB and BA are not equal for the matrices

An Example in Which Cancellation Is Not Valid

Show that AC = BC.

For instance, for n = 1, 2, and 3,

Multiplication by an Identity Matrix

It is convenient also to define A0 = In (where A is a square matrix of order n). These

Repeated Multiplication of a Square Matrix

Theorem 2.5 Number of Solutions of a Linear System

the trAnsPOse Of A MAtrix

Find the transpose of each matrix.

theOreM 2.6 Properties of transposes

See LarsonLinearAlgebra.com for an interactive version of this type of example.

Show that (AB)T and BTAT are equal.

the Product of a Matrix and its transpose

Information retrieval systems such as Internet search engines

2.2 Exercises See CalcChat.com for worked-out solutions to odd-numbered exercises.

Evaluating an Expression In Exercises 1–6, evaluate Associativity of Matrix Multiplication In Exercises 23

13. Solve for X in the equation, given

59. f (x) = 2 − 5x + x2, A= [24 0

2.3 The Inverse of a Matrix

Matrices and Their Inverses

Definition of the Inverse of a Matrix

Theorem 2.7 Uniqueness of an Inverse Matrix

The Inverse of a Matrix

Show that B is the inverse of A, where

Finding the Inverse of a Matrix

Find the inverse of the matrix

Generalizing the method used to solve Example 2 provides a convenient method

finding the inverse of a Matrix by Gauss-Jordan elimination

Recall Hooke’s law, which states that for relatively small

finding the inverse of a Matrix

See LarsonLinearAlgebra.com for an interactive version of this type of example.

Find the inverse of the matrix.

Show that the matrix has no inverse.

If possible, find the inverse of each matrix.

 or the matrix B, you have ad − bc = (3)(2) − (−1)(−6) = 0, which means that

theoreM 2.8 properties of inverse Matrices

(cA) (1cA ) = (c1c )AA

Theorem 2.1 Properties of Matrix Addition

or the matrix B, you have ad − bc = (3)(2) − (−1)(−6) = 0, which means that

Solving Systems of Equations

Elementary Matrices and

Elementary Matrices and

Writing a Matrix as the Product

Examples of Stochastic Matrices

Finding Steady State Matrices of Absorbing