26201_02_ch02_p023-047.

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2 MATRIX ALGEBRA

2.1 Definition of a Matrix

2.2 Types of Matrices
2.3 Matrix Operations
2.4 Gauss–Jordan Elimination Method
Summary
Problems

Somerset Corporate Center Office Building, New Jersey, and its Analytical Model
(Photo courtesy of Ram International. Structural Engineer: The Cantor Seinuk Group, P.C.)

23

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24 Chapter 2 Matrix Algebra

In matrix methods of structural analysis, the fundamental relationships of

equilibrium, compatibility, and member force–displacement relations are
expressed in the form of matrix equations, and the analytical procedures are
formulated by applying various matrix operations. Therefore, familiarity
with the basic concepts of matrix algebra is a prerequisite to understanding
matrix structural analysis. The objective of this chapter is to concisely pre-
sent the basic concepts of matrix algebra necessary for formulating the
methods of structural analysis covered in the text. A general procedure for
solving simultaneous linear equations, the Gauss–Jordan method, is also
discussed.
We begin with the basic definition of a matrix in Section 2.1, followed by
brief descriptions of the various types of matrices in Section 2.2. The matrix
operations of equality, addition and subtraction, multiplication, transposition,
differentiation and integration, inversion, and partitioning are defined in Sec-
tion 2.3; we conclude the chapter with a discussion of the Gauss–Jordan elim-
ination method for solving simultaneous equations (Section 2.4).

2.1 DEFINITION OF A MATRIX

A matrix is defined as a rectangular array of quantities arranged in rows
and columns. A matrix with m rows and n columns can be expressed as
follows.
⎡ ⎤
A11 A12 A13 · · · · · · A1n
⎢ A21 A22 A23 · · · · · · A2n ⎥
⎢ ⎥
⎢
A = [A] = ⎢ A31 A32 A33 · · · · · · A3n ⎥ (2.1)
⎥
⎣ ··· ··· · · · · · · Ai j · · · ⎦ ith row
Am1 Am2 Am3 · · · · · · Amn
jth column m×n

As shown in Eq. (2.1), matrices are denoted either by boldface letters (A) or
by italic letters enclosed within brackets ([A]). The quantities forming a
matrix are referred to as its elements. The elements of a matrix are usually
numbers, but they can be symbols, equations, or even other matrices (called
submatrices). Each element of a matrix is represented by a double-subscripted
letter, with the first subscript identifying the row and the second subscript
identifying the column in which the element is located. Thus, in Eq. (2.1),
A23 represents the element located in the second row and third column of
matrix A. In general, Aij refers to an element located in the ith row and jth
column of matrix A.
The size of a matrix is measured by the number of its rows and columns
and is referred to as the order of the matrix. Thus, matrix A in Eq. (2.1), which
has m rows and n columns, is considered to be of order m × n (m by n). As an

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Section 2.2 Types of Matrices 25

example, consider a matrix D given by

⎡ ⎤
3 5 37
⎢ 8 −6 0⎥
D=⎢ ⎣ 12 23
⎥
2⎦
7 −9 −1
The order of this matrix is 4 × 3, and its elements are symbolically denoted
by Dij with i = 1 to 4 and j = 1 to 3; for example, D13 = 37, D31 = 12,
D42 = −9, etc.

2.2 TYPES OF MATRICES

We describe some of the common types of matrices in the following paragraphs.

Column Matrix (Vector)

If all the elements of a matrix are arranged in a single column (i.e., n = 1), it is
called a column matrix. Column matrices are usually referred to as vectors, and
are sometimes denoted by italic letters enclosed within braces. An example of
a column matrix or vector is given by
⎡ ⎤
35
⎢ 9⎥
⎢ ⎥
B = {B} = ⎢ ⎢ 12 ⎥
⎥
⎣ 3⎦
26

Row Matrix
A matrix with all of its elements arranged in a single row (i.e., m = 1) is re-
ferred to as a row matrix. For example,
C = [9 35 −12 7 22]

Square Matrix
If a matrix has the same number of rows and columns (i.e., m = n), it is called
a square matrix. An example of a 4 × 4 square matrix is given by
⎡ ⎤
⎢ 6 12 0 20 ⎥
⎢ 15 −9 −37 ⎥
⎢ 3 ⎥
A=⎢ ⎥ (2.2)
⎢ −24 13 8 1 ⎥
⎣ ⎦
40 0 11 −5

main diagonal

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26 Chapter 2 Matrix Algebra

As shown in Eq. (2.2), the main diagonal of a square matrix extends from the
upper left corner to the lower right corner, and it contains elements with match-
ing subscripts—that is, A11, A22, A33, . . . , Ann. The elements forming the main
diagonal are referred to as the diagonal elements; the remaining elements of a
square matrix are called the off-diagonal elements.

Symmetric Matrix
When the elements of a square matrix are symmetric about its main diagonal
(i.e., Aij = Aji), it is termed a symmetric matrix. For example,
⎡ ⎤
6 15 −24 40
⎢ 15 −9 13 0⎥
A=⎢
⎣ −24 13
⎥
8 11 ⎦
40 0 11 −5

Lower Triangular Matrix

If all the elements of a square matrix above its main diagonal are zero, (i.e.,
Aij = 0 for j > i), it is referred to as a lower triangular matrix. An example of
a 4 × 4 lower triangular matrix is given by
⎡ ⎤
8 0 0 0
⎢ 12 −9 0 0⎥
A=⎢
⎣ 33
⎥
17 6 0⎦
−2 5 15 3

Upper Triangular Matrix

When all the elements of a square matrix below its main diagonal are zero (i.e.,
Aij = 0 for j < i), it is called an upper triangular matrix. An example of a 3 × 3
upper triangular matrix is given by
⎡ ⎤
−7 6 17
A = ⎣ 0 12 11 ⎦
0 0 20

Diagonal Matrix
A square matrix with all of its off-diagonal elements equal to zero (i.e., Aij = 0
for i  j ), is called a diagonal matrix. For example,
⎡ ⎤
6 0 0 0
⎢0 −3 0 0⎥
A=⎢
⎣0
⎥
0 11 0⎦
0 0 0 27

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Section 2.3 Matrix Operations 27

Unit or Identity Matrix

If all the diagonal elements of a diagonal matrix are equal to 1 (i.e., Iij = 1 and
Iij = 0 for i = j), it is referred to as a unit (or identity) matrix. Unit matrices are
commonly denoted by I or [I]. An example of a 3 × 3 unit matrix is given by
⎡ ⎤
1 0 0
I = ⎣0 1 0⎦
0 0 1

Null Matrix
If all the elements of a matrix are zero (i.e., Oij = 0), it is termed a null matrix.
Null matrices are usually denoted by O or [O]. An example of a 3 × 4 null
matrix is given by
⎡ ⎤
0 0 0 0
O = ⎣0 0 0 0⎦
0 0 0 0

2.3 MATRIX OPERATIONS

Equality
Matrices A and B are considered to be equal if they are of the same order and if
their corresponding elements are identical (i.e., Aij = Bij). Consider, for
example, matrices
⎡ ⎤ ⎡ ⎤
6 2 6 2
A = ⎣ −7 8⎦ and B = ⎣ −7 8⎦
3 −9 3 −9
Since both A and B are of order 3 × 2, and since each element of A is equal to
the corresponding element of B, the matrices A and B are equal to each other;
that is, A = B.

Addition and Subtraction

Matrices can be added (or subtracted) only if they are of the same order. The
addition (or subtraction) of two matrices A and B is carried out by adding
(or subtracting) the corresponding elements of the two matrices. Thus, if
A + B = C, then Cij = Aij + Bij; and if A − B = D, then Dij = Aij − Bij . The
matrices C and D have the same order as matrices A and B.

EXAMPLE 2.1 Calculate the matrices C = A + B and D = A − B if

⎡ ⎤ ⎡ ⎤
6 0 2 3
A = ⎣ −2 9 ⎦ and B=⎣ 7 5⎦
5 1 −12 −1

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28 Chapter 2 Matrix Algebra

SOLUTION
⎡ ⎤ ⎡ ⎤
(6 + 2) (0 + 3) 8 3
C = A + B = ⎣ (−2 + 7) (9 + 5) ⎦ = ⎣ 5 14 ⎦ Ans
(5 − 12) (1 − 1) −7 0
⎡ ⎤ ⎡ ⎤
(6 − 2) (0 − 3) 4 −3
D = A − B = ⎣ (−2 − 7) (9 − 5) ⎦ = ⎣ −9 4⎦ Ans
(5 + 12) (1 + 1) 17 2

Multiplication by a Scalar
The product of a scalar c and a matrix A is obtained by multiplying each
element of the matrix A by the scalar c. Thus, if cA = B, then Bij = cAij .

EXAMPLE 2.2 Calculate the matrix B = cA if c = −6 and

⎡ ⎤
3 7 −2
A=⎣ 0 8 1⎦
12 −4 10

SOLUTION
⎡ ⎤ ⎡ ⎤
−6(3) −6(7) −6(−2) −18 −42 12
B = cA = ⎣ −6(0) −6(8) −6(1) ⎦ = ⎣ 0 −48 −6 ⎦ Ans
−6(12) −6(−4) −6(10) −72 24 −60

Multiplication of Matrices
Two matrices can be multiplied only if the number of columns of the first ma-
trix equals the number of rows of the second matrix. Such matrices are said to
be conformable for multiplication. Consider, for example, the matrices
⎡ ⎤
1 8 
6 −7
A=⎣ 4 −2 ⎦ and B= (2.3)
−1 2
−5 3
3×2 2×2

The product AB of these matrices is defined because the first matrix, A, of the
sequence AB has two columns and the second matrix, B, has two rows. How-
ever, if the sequence of the matrices is reversed, then the product BA does not
exist, because now the first matrix, B, has two columns and the second matrix,
A, has three rows. The product AB is referred to either as A postmultiplied by
B, or as B premultiplied by A. Conversely, the product BA is referred to either
as B postmultiplied by A, or as A premultiplied by B.
When two conformable matrices are multiplied, the product matrix thus
obtained has the number of rows of the first matrix and the number of columns

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Section 2.3 Matrix Operations 29

of the second matrix. Thus, if a matrix A of order l × m is postmultiplied by a

matrix B of order m × n, then the product matrix C = AB has the order l × n;
that is,

A B = C
(l × m) (m × n) (l × n)
equal
(2.4)

⎡ ⎤ ⎡ ⎤
⎡ ⎤
⎢ ⎥ B1 j ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥
i th row ⎢
⎢ Ai1 Ai2 ··· ··· Aim ⎥
⎥⎢
⎢ B2 j ⎥ ⎢
⎥ ⎢ Ci j ⎥ i th row
⎥
..
⎢ ⎥⎢ . ⎥=⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ .. ⎥ ⎢ ⎥
⎢ ⎥⎣ . ⎦ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
Bm j

j th column
j th column

Any element Cij of the product matrix C can be determined by multiplying

each element of the ith row of A by the corresponding element of the jth
column of B (see Eq. 2.4), and by algebraically summing the products; that is,
Ci j = Ai1 B1 j + Ai2 B2 j + · · · + Aim Bm j (2.5)
Eq. (2.5) can be expressed as

m
Ci j = Aik Bk j (2.6)
k=1

in which m represents the number of columns of A, or the number of rows

of B. Equation (2.6) can be used to determine all elements of the product
matrix C = AB.

EXAMPLE 2.3 Calculate the product C = AB of the matrices A and B given in Eq. (2.3).

SOLUTION
⎡ ⎤ ⎡ ⎤
1 8  −2 9
6 −7
C = AB = ⎣ 4 −2 ⎦ = ⎣ 26 −32 ⎦ Ans
−1 2
−5 3 −33 41
(3 × 2) (2 × 2) (3 × 2)
The element C11 of the product matrix C is determined by multiplying each element of
the first row of A by the corresponding element of the first column of B and summing

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30 Chapter 2 Matrix Algebra

the resulting products; that is,

C11 = 1(6) + 8(−1) = −2
Similarly, the element C12 is obtained by multiplying the elements of the first row of
A by the corresponding elements of the second column of B and adding the resulting
products; that is,
C12 = 1(−7) + 8(2) = 9
The remaining elements of C are computed in a similar manner:
C21 = 4(6) + (−2)(−1) = 26
C22 = 4(−7) −2(2) = −32
C31 = −5(6) + 3(−1) = −33
C32 = −5(−7) + 3(2) = 41

A flowchart for programming the matrix multiplication procedure on a com-

puter is given in Fig. 2.1. Any programming language (such as FORTRAN,
BASIC, or C, among others) can be used for this purpose. The reader is encour-
aged to write this program in a general form (e.g., as a subroutine), so that it can
be included in the structural analysis computer programs to be developed in later
chapters.
An important application of matrix multiplication is to express simultane-
ous equations in compact matrix form. Consider the following system of linear
simultaneous equations.
A11 x1 + A12 x2 + A13 x3 + A14 x4 = P1
A21 x1 + A22 x2 + A23 x3 + A24 x4 = P2
(2.7)
A31 x1 + A32 x2 + A33 x3 + A34 x4 = P3
A41 x1 + A42 x2 + A43 x3 + A44 x4 = P4
in which xs are the unknowns and As and Ps represent the coefficients and con-
stants, respectively. By using the definition of multiplication of matrices, this
system of equations can be expressed in matrix form as
⎡ ⎤⎡ ⎤ ⎡ ⎤
A11 A12 A13 A14 x1 P1
⎢A ⎥ ⎢ ⎥ ⎢ ⎥
⎢ 21 A22 A23 A24 ⎥ ⎢ 2 ⎥ ⎢ P2 ⎥
x
(2.8)
⎢A ⎥⎢ ⎥ = ⎢ ⎥
⎣ 31 A32 A33 A34 ⎦ ⎣ x3 ⎦ ⎣ P3 ⎦
A41 A42 A43 A44 x4 P4

or, symbolically, as
Ax = P (2.9)
Matrix multiplication is generally not commutative; that is,

AB = BA (2.10)

Even when the orders of two matrices A and B are such that both products AB
and BA are defined and are of the same order, the two products, in general, will

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Section 2.3 Matrix Operations 31

Start

Input A(L, M), B(M, N)

Dimension C(L, N)

I=1

no
I ≤ L?

yes
J=1

no
J ≤ N?

yes

C(I, J) = 0.0

K=1

no
K ≤ M?

yes
C(I, J) = C(I, J) + A(I, K)*B(K, J)

K=K+1

J=J+1

I=I+1

Output C

Stop

Fig. 2.1 Flowchart for Matrix Multiplication

not be equal. It is essential, therefore, to maintain the proper sequential order

of matrices when evaluating matrix products.

EXAMPLE 2.4 Calculate the products AB and BA if

 
1 −8 6 −3
A= and B=
−7 2 4 −5
Are the products AB and BA equal?

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32 Chapter 2 Matrix Algebra

SOLUTION
  
1 −8 6 −3 −26 37
AB = = Ans
−7 2 4 −5 −34 11
  
6 −3 1 −8 27 −54
BA = = Ans
4 −5 −7 2 39 −42
Comparing products AB and BA, we can see that AB = BA. Ans

Matrix multiplication is associative and distributive, provided that the se-

quential order in which the matrices are to be multiplied is maintained. Thus,
ABC = (AB)C = A(BC) (2.11)
and
A(B + C) = AB + AC (2.12)
The product of any matrix A and a conformable null matrix O equals a
null matrix; that is,
AO = O and OA = O (2.13)
For example,
  
2 −4 0 0 0 0
=
−6 8 0 0 0 0
The product of any matrix A and a conformable unit matrix I equals the
original matrix A; thus,
AI = A and IA = A (2.14)
For example,
  
2 −4 1 0 2 −4
=
−6 8 0 1 −6 8
and
  
1 0 2 −4 2 −4
=
0 1 −6 8 −6 8
We can see from Eqs. (2.13) and (2.14) that the null and unit matrices serve
purposes in matrix algebra that are similar to those of the numbers 0 and 1, re-
spectively, in scalar algebra.

Transpose of a Matrix
The transpose of a matrix is obtained by interchanging its corresponding rows
and columns. The transposed matrix is commonly identified by placing a
superscript T on the symbol of the original matrix. Consider, for example, a
3 × 2 matrix
⎡ ⎤
2 −4
B = ⎣ −5 8⎦
1 3
3×2

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Section 2.3 Matrix Operations 33

The transpose of B is given by


2 −5 1
B =
T
−4 8 3
2×3
Note that the first row of B becomes the first column of BT. Similarly, the sec-
ond and third rows of B become, respectively, the second and third columns of
BT. The order of BT thus obtained is 2 × 3.
As another example, consider the matrix
⎡ ⎤
2 −1 6
C = ⎣ −1 7 −9 ⎦
6 −9 5

Because the elements of C are symmetric about its main diagonal (i.e.,
Cij = Cji for i  j), interchanging the rows and columns of this matrix
produces a matrix CT that is identical to C itself; that is, CT = C. Thus, the
transpose of a symmetric matrix equals the original matrix.
Another useful property of matrix transposition is that the transpose of a
product of matrices equals the product of the transposed matrices in reverse
order. Thus,

(AB)T = BTAT (2.15)

Similarly,
(ABC)T = CTBTAT (2.16)

EXAMPLE 2.5 Show that (AB)T = BTAT if

⎡ ⎤
9 −5 
6 −1 10
A=⎣ 2 1⎦ and B=
−2 7 5
−3 4

SOLUTION
⎡⎤ ⎡ ⎤
9 −5  64 −44 65
6 −1 10
AB = ⎣ 2 1⎦ = ⎣ 10 5 25 ⎦
−2 7 5
−3 4 −26 31 −10
⎡ ⎤
64 10 −26
(AB)T = ⎣ −44 5 31 ⎦ (1)
65 25 −10
⎡ ⎤ ⎡ ⎤
6 −2  64 10 −26
9 2 −3
B A = ⎣ −1
T T
7⎦ = ⎣ −44 5 31 ⎦ (2)
−5 1 4
10 5 65 25 −10
By comparing Eqs. (1) and (2), we can see that (AB)T = BT AT . Ans

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34 Chapter 2 Matrix Algebra

Differentiation and Integration

A matrix can be differentiated (or integrated) by differentiating (or integrating)
each of its elements.

EXAMPLE 2.6 Determine the derivative dA/dx if

⎡ ⎤
x2 3 sin x −x 4
A = ⎣ 3 sin x −x cos2 x ⎦
−x 4 2
cos x 7x 3
SOLUTION By differentiating the elements of A, we obtain
d A11
A11 = x 2 = 2x
dx
d A21 d A12
A21 = A12 = 3 sin x = = 3 cos x
dx dx
d A31 d A13
A31 = A13 = −x 4 = = −4x 3
dx dx
d A22
A22 = −x = −1
dx
d A32 d A23
A32 = A23 = cos2 x = = −2 cos x sin x
dx dx
d A33
A33 = 7x 3 = 21x 2
dx
Thus, the derivative dA/dx is given by
⎡ ⎤
2x 3 cos x −4x 3
dA ⎣
= 3 cos x −1 −2 cos x sin x ⎦ Ans
dx
−4x 3 −2 cos x sin x 21x 2

EXAMPLE 2.7 Determine the partial derivative ∂B/∂y if

⎡ 3 ⎤
2y −yz −2x z
B = ⎣ 3x y 2 yz −z 2 ⎦
2x 2 −2x z 3x y 2
SOLUTION We determine the partial derivative, ∂Bij /∂y, of each element of B to obtain
⎡ 2 ⎤
6y −z 0
∂B ⎣
= 6x y z 0 ⎦ Ans
∂y
0 0 6x y

L
EXAMPLE 2.8 Calculate the integral 0 AAT d x if
⎡ ⎤
x
⎢ 1 −
L⎥
A=⎢ ⎣ x ⎦
⎥

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Section 2.3 Matrix Operations 35

SOLUTION First, we calculate the matrix product AAT as

⎡  ⎤
⎡ x ⎤ x 2 x  x
1−   1 − 1−
⎢
B = AAT = ⎣ L ⎥ 1− x x =⎢ ⎢ L L L ⎥⎥
⎦ ⎣x 
x L L x x2 ⎦
L 1 −
L L L2

Next, we integrate the elements of B to obtain

 L  L  L 
x 2 2x x2
B11 d x = 1− dx = 1− + 2 dx
0 0 L 0 L L
  L
x2 x3 L
= x− + 2
=
L 3L 0 3
 L  L  L   L 
x x x x2
B21 d x = B12 d x = 1− dx = − 2 dx
0 0 0 L L 0 L L
 2  L
x x3 L L L
= − 2
= − =
2L 3L 0 2 3 6
 L  L 2  3 L
x x L
B22 d x = dx = =
0 0 L2 3L 2 0 3

Thus,
⎡L L⎤
 L 
AAT d x = ⎣ 3 6 ⎦= L 2 1
Ans
0 L L 6 1 2
6 3

Inverse of a Square Matrix

The inverse of a square matrix A is defined as a matrix A−1 with elements of
such magnitudes that the product of the original matrix A and its inverse A1
equals a unit matrix I; that is,

AA−1 = A−1 A = I (2.17)

The operation of inversion is defined only for square matrices, with the inverse
of such a matrix also being a square matrix of the same order as the original
matrix. A procedure for determining inverses of matrices will be presented in
the next section.

EXAMPLE 2.9 Check whether or not matrix B is the inverse of matrix A, if

 
−4 2 0.5 −1
A= and B=
−3 1 1.5 −2

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36 Chapter 2 Matrix Algebra

SOLUTION
   
−4 2 0.5 −1 (−2 + 3) (4 − 4) 1 0
AB = = =
−3 1 1.5 −2 (−1.5 + 1.5) (3 − 2) 0 1

Also,
   
0.5 −1 −4 2 (−2 + 3) (1 − 1) 1 0
BA = = =
1.5 −2 −3 1 (−6 + 6) (3 − 2) 0 1
Since AB = BA = I, B is the inverse of A; that is,
B = A−1 Ans

The operation of matrix inversion serves a purpose analogous to the oper-

ation of division in scalar algebra. Consider a system of simultaneous linear
equations expressed in matrix form as
Ax = P
in which A is the square matrix of known coefficients; x is the vector of the
unknowns; and P is the vector of the constants. As the operation of division is
not defined in matrix algebra, the equation cannot be solved for x by dividing
P by A (i.e., x = P/A). However, we can determine x by premultiplying both
sides of the equation by A−1, to obtain
A−1Ax = A−1P
As A−1A = I and Ix = x, we can write
x = A−1P
which shows that a system of simultaneous linear equations can be solved
by premultiplying the vector of constants by the inverse of the coefficient
matrix.
An important property of matrix inversion is that the inverse of a symmet-
ric matrix is also a symmetric matrix.

Orthogonal Matrix
If the inverse of a matrix is equal to its transpose, the matrix is referred to as an
orthogonal matrix. In other words, a matrix A is orthogonal if
A−1 = AT

EXAMPLE 2.10 Determine whether matrix A given below is an orthogonal matrix.

⎡ ⎤
0.8 0.6 0 0
⎢ −0.6 0.8 0 0 ⎥
A=⎢ ⎣ 0
⎥
0 0.8 0.6 ⎦
0 0 −0.6 0.8

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Section 2.3 Matrix Operations 37

SOLUTION
⎡ ⎤⎡ ⎤
0.8 0.6 0 0 0.8 −0.6 0 0
⎢ −0.6 0.8 0 0 ⎥ ⎢ 0 ⎥
AAT = ⎢ ⎥ ⎢ 0.6 0.8 0 ⎥
⎣ 0 0 0.8 0.6 ⎦ ⎣ 0 0 0.8 −0.6 ⎦
0 0 −0.6 0.8 0 0 0.6 0.8
⎡ ⎤
(0.64 + 0.36) (−0.48 + 0.48) 0 0
⎢ (−0.48 + 0.48) (0.36 + 0.64) 0 0 ⎥
=⎢⎣
⎥
0 0 (0.64 + 0.36) (−0.48 + 0.48) ⎦
0 0 (−0.48 + 0.48) (0.36 + 0.64)
⎡ ⎤
1 0 0 0
⎢0 1 0 0⎥
=⎢⎣0 0 1 0⎦
⎥

0 0 0 1
which shows that AAT = I. Thus,
A−1 = AT
Therefore, matrix A is orthogonal. Ans

Partitioning of Matrices
In many applications, it becomes necessary to subdivide a matrix into a num-
ber of smaller matrices called submatrices. The process of subdividing a matrix
into submatrices is referred to as partitioning. For example, a 4 × 3 matrix B
is partitioned into four submatrices by drawing horizontal and vertical dashed
partition lines:
⎡ ⎤
2 −4 −1 
⎢ −5 
⎢ 7 3 ⎥ ⎥ B11 B12
B=⎣ = (2.18)
8 −9 6 ⎦ B21 B22
  
1 3 8
in which the submatrices are
⎡ ⎤ ⎡ ⎤
2 −4 −1
B11 = ⎣ −5 7⎦ B12 = ⎣ 3 ⎦
8 −9 6
B21 = [1 3] B22 = [8]
Matrix operations (such as addition, subtraction, and multiplication) can be
performed on partitioned matrices in the same manner as discussed previously
by treating the submatrices as elements—provided that the matrices are parti-
tioned in such a way that their submatrices are conformable for the particular op-
eration. For example, suppose that the 4 × 3 matrix B of Eq. (2.18) is to be post-
multiplied by a 3 × 2 matrix C, which is partitioned into two submatrices:
⎡ ⎤
9 −6 
C11
C=⎣ 4 2⎦ = (2.19)
 C21
−3 1

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38 Chapter 2 Matrix Algebra

The product BC is expressed in terms of submatrices as

  
B11 B12 C11 B11 C11 + B12 C21
BC = = (2.20)
B21 B22 C21 B21 C11 + B22 C21

It is important to realize that matrices B and C have been partitioned in such a

way that their corresponding submatrices are conformable for multiplication;
that is, the orders of the submatrices are such that the products B11C11, B12C21,
B21C11, and B22C21 are defined. It can be seen from Eqs. (2.18) and (2.19) that
this is achieved by partitioning the rows of the second matrix C of the product
BC in the same way that the columns of the first matrix B are partitioned. The
products of the submatrices are:
⎡ ⎤ ⎡ ⎤
2 −4  2 −20
9 −6
B11 C11 = ⎣ −5 7⎦ = ⎣ −17 44 ⎦
4 2
8 −9 36 −66
⎡ ⎤ ⎡ ⎤
−1 3 −1
B12 C21 = ⎣ 3 ⎦ [−3 1] = ⎣ −9 3⎦
6 −18 6

9 −6
B21 C11 = [1 3] = [21 0]
4 2
B22 C21 = [8][−3 1] = [−24 8]
By substituting the numerical values of the products of submatrices into
Eq. (2.20), we obtain
⎡⎡ ⎤ ⎡ ⎤⎤ ⎡ ⎤
2 −20 3 −1 5 −21
⎢ ⎣ −17 44 ⎦ + ⎣ −9 3⎦⎥ ⎢ 47 ⎥
BC = ⎢ ⎥ = ⎢ −26 ⎥
⎣ 36 −66 −18 6 ⎦ ⎣ 18 −60 ⎦
[21 0] + [−24 8] −3 8

2.4 GAUSS–JORDAN ELIMINATION METHOD

The Gauss–Jordan elimination method is one of the most commonly used
procedures for solving simultaneous linear equations, and for determining
inverses of matrices.

Solution of Simultaneous Equations

To illustrate the Gauss–Jordan method for solving simultaneous equations,
consider the following system of three linear algebraic equations:
5x1 + 6x2 − 3x3 = 66
9x1 − x2 + 2x3 = 8 (2.21a)
8x1 − 7x2 + 4x3 = −39

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Section 2.4 Gauss–Jordan Elimination Method 39

To determine the unknowns x1, x2, and x3, we begin by dividing the first equa-
tion by the coefficient of its x1 term to obtain
x1 + 1.2x2 − 0.6x3 = 13.2
9x1 − x2 + 2x3 = 8 (2.21b)
8x1 − 7x2 + 4x3 = −39
Next, we eliminate the unknown x1 from the second and third equations by suc-
cessively subtracting from each equation the product of the coefficient of its x1
term and the first equation. Thus, to eliminate x1 from the second equation, we
multiply the first equation by 9 and subtract it from the second equation.
Similarly, we eliminate x1 from the third equation by multiplying the first equa-
tion by 8 and subtracting it from the third equation. This yields the system of
equations
x1 + 1.2x2 − 0.6x3 = 13.2
− 11.8x2 + 7.4x3 = −110.8 (2.21c)
− 16.6x2 + 8.8x3 = −144.6
With x1 eliminated from all but the first equation, we now divide the second
equation by the coefficient of its x2 term to obtain
x1 + 1.2x2 − 0.6 x3 = 13.2
x2 − 0.6271x3 = 9.39 (2.21d)
− 16.6x2 + 8.8 x3 = −144.6
Next, the unknown x2 is eliminated from the first and the third equations, suc-
cessively, by multiplying the second equation by 1.2 and subtracting it from the
first equation, and then by multiplying the second equation by −16.6 and sub-
tracting it from the third equation. The system of equations thus obtained is
x1 + 0.1525x3 = 1.932
x2 − 0.6271x3 = 9.39 (2.21e)
− 1.61 x3 = 11.27
Focusing our attention now on the unknown x3, we divide the third equation by
the coefficient of its x3 term (which is −1.61) to obtain
x1 + 0.1525x3 = 1.932
x2 − 0.6271x3 = 9.39 (2.21f)
x3 = −7
Finally, we eliminate x3 from the first and the second equations, successively,
by multiplying the third equation by 0.1525 and subtracting it from the first
equation, and then by multiplying the third equation by −0.6271 and subtract-
ing it from the second equation. This yields the solution of the given system of
equations:
x1 = 3
x2 = 5 (2.21g)
x3 = −7

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40 Chapter 2 Matrix Algebra

or, equivalently,
x1 = 3; x2 = 5; x3 = −7 (2.21h)
To check that this solution is correct, we substitute the numerical values of
x1, x2, and x3 back into the original equations (Eq. 2.21(a)):
5(3) + 6(5) − 3(−7) = 66 Checks

9(3) − 5 + 2(−7) = 8 Checks

8(3) − 7(5) + 4(−7) = −39 Checks

As the foregoing example illustrates, the Gauss–Jordan method basically

involves eliminating, in order, each unknown from all but one of the equations
of the system by applying the following operations: dividing an equation by a
scalar; and multiplying an equation by a scalar and subtracting the resulting
equation from another equation. These operations (called the elementary
operations) when applied to a system of equations yield another system of
equations that has the same solution as the original system. In the Gauss–-
Jordan method, the elementary operations are performed repeatedly until a
system with each equation containing only one unknown is obtained.
The Gauss–Jordan elimination method can be performed more conve-
niently by using the matrix form of the simultaneous equations (Ax = P). In
this approach, the coefficient matrix A and the vector of constants P are treated
as submatrices of a partitioned augmented matrix,
G = [A  P]
(2.22)
n × (n + 1) n × n n×1
where n represents the number of equations. The elementary operations are
then applied to the rows of the augmented matrix, until the coefficient matrix is
reduced to a unit matrix. The elements of the vector, which initially contained
the constant terms of the original equations, now represent the solution of the
original system of equations; that is,
⎧
⎨ [A  P] 
G= −−−−−−− elementary operations (2.23)
⎩
[I  x]
This procedure is illustrated by the following example.

EXAMPLE 2.11 Solve the system of simultaneous equations given in Eq. 2.21(a) by the Gauss–Jordan
method.

SOLUTION The given system of equations can be written in matrix form as

Ax = P
⎡ ⎤⎡ ⎤ ⎡ ⎤
5 6 −3 x1 66
⎣ 9 −1 2 ⎦ ⎣ x2 ⎦ = ⎣ 8 ⎦ (1)
8 −7 4 x3 −39

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Section 2.4 Gauss–Jordan Elimination Method 41

from which we form the augmented matrix

⎡ ⎤
5 6 −3 66
G = [AP] = ⎣ 9 −1 
2 8 ⎦ (2)
8 −7 4−39
We begin Gauss–Jordan elimination by dividing row 1 of the augmented matrix by
G11 = 5 to obtain
⎡ ⎤
1 1.2 −0.6 13.2
G = ⎣ 9 −1 
2  8 ⎦ (3)
8 −7 4 −39
Next, we multiply row 1 by G21 = 9 and subtract it from row 2; then multiply row 1
by G31 = 8 and subtract it from row 3. This yields
⎡ ⎤
1 1.2 −0.6  13.2
G = ⎣ 0 −11.8 7.4 
−110.8
⎦ (4)
0 −16.6 
8.8 −144.6
We now divide row 2 by G22 = −11.8 to obtain
⎡ ⎤
1 1.2 −0.6  13.2
G = ⎣0 1 −0.6271
 9.39 ⎦ (5)
0 −16.6 8.8 −144.6
Next, we multiply row 2 by G12 = 1.2 and subtract it from row 1, and then multiply
row 2 by G32 = −16.6 and subtract it from row 3. Thus,
⎡ ⎤
1 0 0.1525 1.932
G = ⎣ 0 1 −0.6271  9.39
⎦ (6)
0 0 −1.61 11.27
By dividing row 3 by G33 = −1.61, we obtain
⎡ ⎤
1 0 0.1525  1.932

G = ⎣ 0 1 −0.6271  9.39 ⎦ (7)
0 0 1 −7
Finally, we multiply row 3 by G13 = 0.1525 and subtract it from row 1; then multiply
row 3 by G23 = −0.6271 and subtract it from row 2 to obtain
⎡ ⎤
1 0 0 3
G = ⎣ 0 1 0  5
⎦ (8)
0 0 1 −7
Thus, the solution of the given system of equations is
⎡ ⎤
3
x = ⎣ 5⎦ Ans

−7
To check our solution, we substitute the numerical value of x back into Eq. (1). This yields
⎡ ⎤⎡ ⎤ ⎡ ⎤
5 6 −3 3 15 + 30 + 21 = 66
⎣ 9 −1 2 ⎦ ⎣ 5 ⎦ = ⎣ 27 − 5 − 14 = 8⎦ Checks
8 −7 4 −7 24 − 35 − 28 = −39

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Start

Input G(N, N + 1) = [A P]

I=1

no
I ≤ N?

yes
C = G(I, I )

J=1

no
J ≤ N + 1?

yes
G(I, J) = G(I, J) / C

J=J+1

K=1

no
K ≤ N?

yes
yes
K = I?

no
D = G(K, I )

M= I

no
M ≤ N + 1?

yes
G(K, M) = G(K, M) − G(I, M)*D

M=M+ 1

K=K+ 1

I=I+ 1

Output G

Stop

Fig. 2.2 Flowchart for Solution of Simultaneous

42 Equations by Gauss–Jordan Method

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Section 2.4 Gauss–Jordan Elimination Method 43

The solution of large systems of simultaneous equations by the Gauss–

Jordan method is usually carried out by computer, and a flowchart for program-
ming this procedure is given in Fig. 2.2. The reader should write this program in a
general form (e.g., as a subroutine), so that it can be conveniently included in the
structural analysis computer programs to be developed in later chapters.
It should be noted that the Gauss–Jordan method as described in the pre-
ceding paragraphs breaks down if a diagonal element of the coefficient matrix
A becomes zero during the elimination process. This situation can be remedied
by interchanging the row of the augmented matrix containing the zero diago-
nal element with another row, to place a nonzero element on the diagonal; the
elimination process is then continued. However, when solving the systems of
equations encountered in structural analysis, the condition of a zero diagonal
element should not arise; the occurrence of such a condition would indicate
that the structure being analyzed is unstable [2]*.

Matrix Inversion
The procedure for determining inverses of matrices by the Gauss–Jordan
method is similar to that described previously for solving simultaneous equa-
tions. The procedure involves forming an augmented matrix G composed of
the matrix A that is to be inverted and a unit matrix I of the same order as A;
that is,
G = [A  I]
(2.24)
n × 2n n × n n×n
Elementary operations are then applied to the rows of the augmented matrix to
reduce A to a unit matrix. Matrix I, which was initially the unit matrix, now
represents the inverse matrix A−1; thus,
⎧
⎨ [A  I]

G= −−−−−−−−− elementary operations (2.25)
⎩
[I  A−1 ]

EXAMPLE 2.12 Determine the inverse of the matrix shown using the Gauss–Jordan method.
⎡ ⎤
13 −6 6
A = ⎣ −6 12 −1 ⎦
6 −1 9

SOLUTION The augmented matrix is given by

⎡ ⎤
13 −6 6 1 0 0
G = [AI] = ⎣ −6 12 −1 0 1 0⎦ (1)
6 −1 9 0 0 1

*Numbers in brackets refer to items listed in the bibliography.

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44 Chapter 2 Matrix Algebra

We begin the Gauss–Jordan elimination process by dividing row 1 of the augmented

matrix by G11 = 13:
⎡ ⎤
1 −0.4615 0.46150.07692 0 0
G = ⎣ −6 12 −1 0 1 0⎦
 (2)
6 −1 9  0 0 1
Next, we multiply row 1 by G21 = −6 and subtract it from row 2, and then multiply
row 1 by G31 = 6 and subtract it from row 3. This yields
⎡ ⎤
1 −0.4615 0.4615 0.07692 0 0
G = ⎣0 9.231 1.769   0.4615 1 0⎦ (3)
0 1.769 6.231 −0.4615 0 1
Dividing row 2 by G22 = 9.231, we obtain
⎡ ⎤
1 −0.4615 0.4615 0.07692 0 0
G = ⎣0 1 0.1916
 0.04999 0.1083 0⎦ (4)
0 1.769 6.231 −0.4615 0 1
Next, we multiply row 2 by G12 = −0.4615 and subtract it from row 1; then multiply
row 2 by G32 = 1.769 and subtract it from row 3. This yields
⎡ ⎤
1 0 0.5499 0.09999 0.04998 0
G = ⎣ 0 1 0.1916  0.04999 0.1083 0⎦ (5)
0 0 5.892 −0.5499 −0.1916 1
Divide row 3 by G33 = 5.892:
⎡ ⎤
1 0 0.5499  0.09999 0.04998 0
G = ⎣ 0 1 0.1916   0.04999 0.1083 0 ⎦ (6)
0 0 1  −0.09333 −0.03252 0.1697
Multiply row 3 by G13 = 0.5499 and subtract it from row 1; then multiply row 3 by
G23 = 0.1916 and subtract it from row 2 to obtain
⎡ ⎤
1 0 0 0.1513 0.06787 −0.09333
G = ⎣ 0 1 0  0.06787 0.1145 −0.03252 ⎦ (7)
0 0 1−0.09333 −0.03252 0.1697
Thus, the inverse of the given matrix A is
⎡ ⎤
0.1513 0.06787 −0.09333
A = ⎣ 0.06787
−1
0.1145 −0.03252 ⎦ Ans
−0.09333 −0.03252 0.1697
Finally, we check our computations by using the relationship AA−1 = I:
⎡ ⎤⎡ ⎤
13 −6 6 0.1513 0.06787 −0.09333
AA = ⎣ −6 12 −1 ⎦ ⎣ 0.06787
−1
0.1145 −0.03252 ⎦
6 −1 9 −0.09333 −0.03252 0.1697
⎡ ⎤
0.9997 0.0002 0
= ⎣0 0.9993 0 ⎦≈I Checks

0 0 0.9998

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Summary 45

SUMMARY
In this chapter, we discussed the basic concepts of matrix algebra that are nec-
essary for formulating the matrix methods of structural analysis:

1. A matrix is defined as a rectangular array of quantities (elements)

arranged in rows and columns. The size of a matrix is measured by its
number of rows and columns, and is referred to as its order.
2. Two matrices are considered to be equal if they are of the same order,
and if their corresponding elements are identical.
3. Two matrices of the same order can be added (or subtracted) by adding
(or subtracting) their corresponding elements.
4. The matrix multiplication AB = C is defined only if the number of
columns of the first matrix A equals the number of rows of the second
matrix B. Any element Cij of the product matrix C can be evaluated by
using the relationship

m
Ci j = Aik Bk j (2.6)
k=1

where m is the number of columns of A, or the number of rows of B.

Matrix multiplication is generally not commutative; that is, AB = BA.
5. The transpose of a matrix is obtained by interchanging its correspond-
ing rows and columns. If C is a symmetric matrix, then CT = C.
Another useful property of matrix transposition is that

(AB)T = BTAT (2.15)

6. A matrix can be differentiated (or integrated) by differentiating (or in-

tegrating) each of its elements.
7. The inverse of a square matrix A is defined as a matrix A−1 which
satisfies the relationship:

AA−1 = A−1A = I (2.17)

8. If the inverse of a matrix equals its transpose, the matrix is called an

orthogonal matrix.
9. The Gauss–Jordan method of solving simultaneous equations essen-
tially involves successively eliminating each unknown from all but one
of the equations of the system by performing the following operations:
dividing an equation by a scalar; and multiplying an equation by a
scalar and subtracting the resulting equation from another equation.
These elementary operations are applied repeatedly until a system with
each equation containing only one unknown is obtained.

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46 Chapter 2 Matrix Algebra

PROBLEMS
Section 2.3

2.1 Determine the matrices C = A + B and D = A − B if 2.9 Show that (ABC)T = CTBTAT by using the following
⎡ ⎤ ⎡ ⎤ matrices
3 8 −1 5 9 −2
⎡ ⎤
A = ⎣ 8 −7 −4 ⎦ B = ⎣ −9 6 3⎦ −9 0

−1 −4 5 2 −3 −4 ⎢ 13 20 ⎥ 15 −1 −4
A=⎢ ⎣ 8 −3 ⎦
⎥ B=
2.2 Determine the matrices C = 2A + B and D = A − 3B if 6 16 9
⎡ ⎤ ⎡ ⎤ −11 −5
8 −6 −3 3 2 −3 ⎡ ⎤
⎢ 1 −2 0⎥ ⎢ −4 3 0⎥ −7 10 6 0
A=⎢ ⎣ −6
⎥ B=⎢ ⎥
C = ⎣ −1 2 −8 −2 ⎦
5 −1 ⎦ ⎣ 2 −8 6⎦
−2 8 0 −1 4 −7 16 12 2 8

2.3 Determine the products C = AB and D = BA if 2.10 Determine the matrix triple product C = BTAB if
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
3 40 −10 −25 5 7 −3
A = [ 4 −6 2 ] B = ⎣ 1⎦ A = ⎣ −10 15 12 ⎦ B = ⎣ −7 8 4⎦
−5 −25 12 30 3 −4 9
2.4 Determine the products C = AB and D = BA if 2.11 Determine the matrix triple product C = BTAB if
⎡ ⎤
4 6 
 300 −100
⎢ −7 −5 ⎥ −1 3 −5 2 A=
A=⎣ ⎢ ⎥ B= −100
1 −9 ⎦
200
−13 −4 7 6 
−3 11 0.6 0.8 −0.6 −0.8
B=
−0.8 0.6 0.8 −0.6
2.5 Determine the products C = AB and D = BA if
2.12 Develop a computer program to determine the matrix
⎡ ⎤ ⎡ ⎤
4 −6 1 3 5 0 triple product C = BTAB, where A is a square matrix of
A = ⎣ −6 5 7⎦ B = ⎣5 7 −2 ⎦ any order. Check the program by solving Problems 2.10
1 7 8 0 −2 9 and 2.11 and comparing the results to those determined by
hand calculations.
2.6 Determine the products C = AB if 2.13 Determine the derivative dA/dx if
⎡ ⎤ ⎡ ⎤
12 −11 10 ⎡ ⎤ −2x 2 3sin x −7x
⎢ 0 13 −1 5 A = ⎣ 3sin x cos2 x −3x 3 ⎦
2 −4 ⎥
A=⎢
⎣ −7
⎥ B = ⎣ 16 −9 0⎦
8⎦ −7x −3x 3sin2 x
3
9
−3 20 −7
6 15 −5
2.14 Determine the derivative d(A + B)/dx if
2.7 Develop a computer program to determine the matrix ⎡ ⎤ ⎡ ⎤
product C = AB of two conformable matrices A and B of −3x 5 2x 2 −x
any order. Check the program by solving Problems 2.4–2.6 ⎢ 4x 2 −x 3 ⎥ ⎢ −12x 8⎥
A=⎢
⎣ −7
⎥ B=⎢ ⎥
and comparing the computer-generated results to those 5x ⎦ ⎣ 2x 3 −3x 2 ⎦
determined by hand calculations. 2x 3 −x 2 −1 6x
2.8 Show that (AB)T = BTAT by using the following matrices
2.15 Determine the derivative d(AB)/dx if
⎡ ⎤
21 10 16 ⎡ ⎤
7 −4 ⎡ ⎤ ⎡ ⎤
⎢ −15 11 0 ⎥ 4x 2 −5x 2 −5x 3 −x
A=⎢ ⎣ 13
⎥ B = ⎣ −1 9⎦
20 −9 ⎦ A=⎣ 2 −3x 3
−x ⎦ B=⎣ 6 −3x ⎦
2
3 −6
7 −17 14 −5x 2 −x 7 2x 2 4x

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26201_02_ch02_p023-047.qxd 12/1/10 4:59 PM Page 47

Problems 47

2.16 Determine the partial derivatives ∂A/∂x, ∂A/∂y, and 2.22 20x1 − 9x2 + 15x3 = 354
∂A/∂z, if − 9x1 + 16x2 − 5x3 = −275
⎡ 2 ⎤
x −y 2 2z 2 15x1 − 5x2 + 18x3 = 307
A = ⎣ −y 2 3x y −yz ⎦
2.23 4x1 − 2x2 + 3x3 = 37.2
2z 2 −yz 4x z
L 3x1 + 5x2 − x3 = −7.2
2.17 Calculate the integral 0 A d x if
⎡ ⎤ x1 − 4x2 + 2x3 = 30.3
−5 −3x 2
⎢ 4x −x 3 ⎥ 2.24 6x1 + 15x2 − 24x3 + 40x4 = 190.9
A=⎢ ⎣ 2x 4
⎥
6⎦ 15x1 + 9x2 − 13x3 = 69.8
5x 2 −x −24x1 − 13x2 + 8x3 − 11x4 = −96.3
L
2.18 Calculate the integral 0 A dx if 40x1 − 11x3 + 5x4 = 119.35
⎡ ⎤ 2x1 − 5x2 + 8x3 + 11x4 =
2x − sin x 2 cos2 x 2.25 39
A = ⎣ − sin x 5 −4x ⎦
3
10x1 + 7x2 + 4x3 − x4 = 127
−4x 3 (1 − x 2 )
2 cos2 x −3x1 + 9x2 + 5x3 − 6x4 = 58
L
2.19 Calculate the integral 0 AB d x if x1 − 4x2 − 2x3 + 9x4 = −14
⎡ ⎤
 −2x x2 2.26 Develop a computer program to solve a system of simulta-
−x 3 2x 2 3
A= B = ⎣ 5 −2x ⎦ neous equations of any size by the Gauss–Jordan method.
2x −x 2 2x 3
3x 3 −3 Check the program by solving Problems 2.21 through
2.25 and comparing the computer-generated results to
2.20 Determine whether the matrices A and B given below are
those determined by hand calculations.
orthogonal matrices.
2.27 through 2.30 Determine the inverse of the matrices shown
⎡ ⎤
−0.28 −0.96 0 0 by the Gauss–Jordan method.
⎢ 0.96 −0.28 0 0 ⎥ ⎡ ⎤
A=⎢ ⎥ 5 3 −4
⎣ 0 −0.28 −0.96 ⎦
0 2.27 A = ⎣ 3 8 −2 ⎦
0 0 0.96 −0.28 −4 −2 7
⎡ ⎤ ⎡ ⎤
−0.28 0.96 0 0 6 −4 1
⎢ 0.96 −0.28 0 ⎥
B=⎢
0 ⎥ 2.28 A = ⎣ −1 9 3⎦
⎣ 0 0 −0.28 0.96 ⎦ 4 2 5
0 0 0.96 −0.28 ⎡ ⎤
7 −6 3 −2
⎢ −6 4 −1 5⎥
Section 2.4 2.29 A = ⎢
⎣ 3 −1
⎥
8 9⎦
2.21 through 2.25 Solve the following systems of simultane- −2 5 9 2
ous equations by the Gauss–Jordan method. ⎡ ⎤
5 −7 −3 11
2.21 2x1 − 3x2 + x3 = −18 ⎢ 10 −6 −13 2⎥
2.30 A = ⎢
⎣ −1 12
⎥
−9x1 + 5x2 + 3x3 = 18 8 −4 ⎦
4x1 + 7x2 − 8x3 = 53 −9 7 −5 6

Copyright 2010 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.