#1: Matrix Operations

1.1 Matrix Addition and Scalar Multiplication

Definition:
Matrix Addition: Add corresponding elements of two matrices of the same size.
Scalar Multiplication: Multiply every element of a matrix by the same scalar value.

Rules:
1. Addition is commutative: A + B = B + A.
2. Addition is associative: A + (B + C) = (A + B) + C .
3. Scalar multiplication distributes over addition:
r(A + B) = rA + rB .
(r + s)A = rA + sA.

Examples:
2×2 Matrices: Let:

1 2 5 6
A=[ ], B=[ ].
3 4 7 8
​ ​ ​ ​

1. Addition:

1+5 2+6 6 8
A+B =[ ]=[ ].
3+7 4+8 10 12
​ ​ ​ ​

2. Scalar Multiplication: Multiply A by 3:

3⋅1 3⋅2 3 6
3A = [ ]=[ ].
3⋅3 3⋅4 9 12
​ ​ ​ ​
3×2 Matrices: Let:

2 1 1 2
C = −1 3 , ​ ​ ​ ​ D = 3 −1 . ​ ​ ​ ​

0 4 5 0

1. Addition:

2+1 1+2 3 3
C + D = −1 + 3 3 + (−1) = 2 2 .
​ ​ ​ ​ ​ ​ ​ ​

0+5 4+0 5 4

2. Scalar Multiplication: Multiply C by −2:

−2 ⋅ 2 −2 ⋅ 1 −4 −2
−2C = −2 ⋅ (−1) −2 ⋅ 3 = 2 −6 .
​ ​ ​ ​ ​ ​ ​ ​

−2 ⋅ 0 −2 ⋅ 4 0 −8

1.2 Matrix Multiplication

Definition:
Multiply the rows of the first matrix by the columns of the second matrix.
AB is defined if the number of columns in A equals the number of rows in B .

Rules:
1. Matrix multiplication is not commutative: AB  BA.
=
2. Matrix multiplication is associative: A(BC) = (AB)C .
3. Distributive over addition: A(B + C) = AB + AC .
Examples:
2×2 Matrices:

2 3 1 0
A=[ ], B=[ ].
1 4 2 1
​ ​ ​ ​

Find AB :

(2)(1) + (3)(2) (2)(0) + (3)(1)

AB = [ ].
(1)(1) + (4)(2) (1)(0) + (4)(1)
​ ​

Simplify:

8 3
AB = [ ].
9 4
​ ​

3×2 and 2×3 Multiplication:

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

5 6

Find AB :

(1)(1) + (2)(2) (1)(0) + (2)(1) (1)(2) + (2)(3)

AB = (3)(1) + (4)(2) (3)(0) + (4)(1) (3)(2) + (4)(3) .
​ ​ ​ ​ ​

(5)(1) + (6)(2) (5)(0) + (6)(1) (5)(2) + (6)(3)

Simplify:

5 2 8
AB = 11 4 18 . ​ ​ ​ ​ ​

17 6 28

#2: Finding the Inverse of a Matrix

2.1 2×2 Inverse

Formula:
a b
For A =[ ], the inverse is:
c d
​ ​

1 d −b
A−1 = [ ], if det(A) = ad − bc =
 0.
ad − bc −c a
​ ​ ​

Example:
Let:

2 1
A=[ ].
3 4
​ ​

1. Compute det(A):

det(A) = (2)(4) − (1)(3) = 8 − 3 = 5.

2. Apply the formula:

1 4 −1
A−1 = [ ].
5 −3 2
​ ​

Simplify:

0.8 −0.2
A−1 = [ ].
−0.6 0.4
​ ​

2.2 3×3 Inverse

Procedure:
1. Form the augmented matrix [A∣I].
2. Use row operations to transform A into I . The right side becomes A−1 .
Example:
Let:

1 2 3
A= 0 1 4 .​ ​ ​ ​

5 6 0

1. Augment:

1 2 3 1 0 0
0 1 4 0 1 0 .
​ ​ ​ ​ ​ ​ ​ ​

5 6 0 0 0 1

2. Perform row operations to reduce A to I . The result is:

−24 18 5
−1
A = 20 −15 −4 .
​ ​ ​ ​ ​

−5 4 1
#3: Conditions for a Matrix to Have an Inverse

Definition:
A square matrix A is invertible if and only if:
1. Determinant Condition: det(A)  0.
=
2. Row-Reduced Form: A can be row-reduced to the identity matrix I .

Key Points:
1. A matrix is singular if det(A) = 0, meaning it has no inverse.
2. A matrix is nonsingular if det(A)  0, meaning it has an inverse.
=

Examples:
2×2 Example:

2 3
A=[ ].
4 6
​ ​

1. Compute det(A):

det(A) = (2)(6) − (3)(4) = 12 − 12 = 0.

Since det(A) = 0, A is singular and does not have an inverse.

3×3 Example:

1 2 3
B= 4 5 6 .
​ ​ ​ ​ ​

7 8 10
 1. Compute det(B) (using the first row for expansion):

5 6 4 6 4 5
det(B) = 1 ⋅ det [ ] − 2 ⋅ det [ ] + 3 ⋅ det [ ].
8 10 7 10 7 8
​ ​ ​ ​ ​ ​

Compute the minors:

5 6
det [ ] = (5)(10) − (6)(8) = 50 − 48 = 2.
8 10
​ ​

4 6
det [ ] = (4)(10) − (6)(7) = 40 − 42 = −2.
7 10
​ ​

4 5
det [ ] = (4)(8) − (5)(7) = 32 − 35 = −3.
7 8
​ ​

Substitute:

det(B) = 1(2) − 2(−2) + 3(−3) = 2 + 4 − 9 = −3.

Since det(B)  0, B is nonsingular and has an inverse.

=

#4: Properties of Matrix Inverses

Key Properties:

1. (A−1 )−1 = A.
2. (AB)−1 = B −1 A−1 .
3. (AT )−1 = (A−1 )T .
4. det(A−1 ) = 1
det(A)
.​

Examples:

Property 1: (A−1 )−1 =A

2×2 Example: Let:
 2 1
A=[ ].
3 4
​ ​

1. Compute A−1 :

1 4 −1
A−1 = [ ].
5 −3 2
​ ​

2. Compute (A−1 )−1 , which should give A:

2 1
(A−1 )−1 = A = [ ].
3 4
​ ​

Property 2: (AB)−1 = B −1 A−1

2×2 Example: Let:

1 2 2 0
A=[ ], B=[ ].
3 4 0 3
​ ​ ​ ​

1. Compute AB :

2 6
AB = [ ].
6 12
​ ​

2. Find (AB)−1 :

1 12 −6
(AB)−1 = [ ].
24 −6 2
​ ​ ​

3. Find A−1 and B −1 :

1 4 −2 0.5 0
A−1 = [ ], B −1 = [ ].
−2 −3 1 0 0.333
​ ​ ​ ​ ​

4. Verify (AB)−1 = B −1 A−1 .

#5: Using the Inverse of a Matrix to Solve Ax =b

Definition:

For a system Ax = b, the solution is:

x = A−1 b,

if A is invertible.

Examples:
2×2 Example: Solve:

2 1 5
A=[ ], b = [ ].
3 4 11
​ ​ ​

1. Compute A−1 :

1 4 −1
A−1 = [ ].
5 −3 2
​ ​

2. Multiply A−1 b:

0.8 −0.2 5
x=[ ][ ].
−0.6 0.4 11
​ ​

Simplify:

2
x = [ ].
1
​

3×3 Example: Solve:

1 2 3 6
A= 0 1 4 ,
​ ​ ​ ​ ​
b = 10 . ​ ​ ​

5 6 0 8

1. Compute A−1 (using Gaussian elimination):

 −24 18 5
−1
A = 20 −15 −4 .
​ ​ ​ ​ ​

−5 4 1

2. Multiply A−1 b:

−24 18 5 6
x = 20 −15 −4
​ ​ ​ ​ ​ 10 .
​ ​ ​

−5 4 1 8

Simplify:

−2
x= 1 . ​ ​ ​

3
#6: Elementary Matrices and Their Role in Row Operations and Inverses

Definition:
An elementary matrix is a matrix obtained by performing a single elementary row operation on the
identity matrix.
1. Row Swap: Interchange two rows of the identity matrix.
2. Row Scaling: Multiply a row by a nonzero scalar.
3. Row Replacement: Replace a row with itself plus a multiple of another row.

Key Points:
1. Inverse of an Elementary Matrix:

The inverse of an elementary matrix corresponds to the inverse of the row operation.
2. Role in Inverses:

Any invertible matrix A can be expressed as a product of elementary matrices.

Examples:
Elementary Matrices:
1. Row Swap (R1 ​ ↔ R2 ): ​

0 1
E=[ ].
1 0
​ ​

2. Row Scaling (R1 ​ → 3R1 ): ​

3 0
E=[ ].
0 1
​ ​

3. Row Replacement (R2 ​

→ R2 − 2R1 ):
​ ​

1 0
E=[ ].
−2 1
​ ​
 2 1
Example: Represent A =[ ] as a product of elementary matrices.
4 3
​ ​

1 0
1. Start with the identity matrix I =[ ].
0 1
​ ​

2. Perform row operations to reduce A to I :

0.5 0
R1 → 12 R1 : E1 = [ ].
0 1
​ ​ ​ ​ ​ ​

1 0
R2 → R2 − 4R1 : E2 = [ ].
−4 1
​ ​ ​ ​ ​ ​

1 0
R2 → 13 R2 : E3 = [ ].
0 0.333
​ ​ ​ ​ ​ ​

3. Write A = E1−1 E2−1 E3−1 .

​ ​ ​

#7: Expressing a Matrix and Its Inverse as a Product of Elementary Matrices

Steps:
1. Reduce A to I using elementary row operations.
2. Each row operation corresponds to an elementary matrix.
3. The matrix A can be expressed as:

A = Ek−1 Ek−1
​
−1
⋯ E1−1 .
​ ​

4. The inverse of A is:

A−1 = E1 E2 ⋯ Ek .
​ ​ ​

Example:
Let:
 1 2
A=[ ].
3 4
​ ​

1. Augment A with I :

1 2 1 0
[ ].
3 4 0 1
​ ​ ​ ​

2. Perform row operations:

R2 → R2 − 3R1 :
​ ​ ​

1 2 1 0
[ ].
0 −2 −3 1
​ ​ ​ ​

1
R2 → ​

−2 R2 :
​ ​

1 2 1 0
[ ].
0 1 1.5 −0.5
​ ​ ​ ​

R1 → R1 − 2R2 :
​ ​ ​

1 0 −2 1
[ ].
0 1 1.5 −0.5
​ ​ ​ ​

3. The right half is A−1 :

−2 1
A−1 = [ ].
1.5 −0.5
​ ​

#8: LU-Factorization

Definition:
LU-factorization expresses A as:

A = LU ,

where L is lower triangular and U is upper triangular.

Steps:
1. Perform Gaussian elimination on A to form U .
2. Construct L using the multipliers from row operations.

Example:
Let:

2 1 1
A = 4 −6 0 .
​ ​ ​ ​ ​

−2 7 2

1. Transform A to U using Gaussian elimination:

R2 → R2 − 2R1 :
​ ​ ​

2 1 1
0 −8 −2 .
​ ​ ​ ​ ​

−2 7 2
R3 → R3 + R1 :
​ ​ ​

2 1 1
0 −8 −2 .
​ ​ ​ ​ ​

0 8 3
2. The resulting U is:

2 1 1
U = 0 −8 −2 . ​ ​ ​ ​ ​

0 0 1
3. Construct L:

1 0 0
L= 2 1 0 . ​ ​ ​ ​ ​

−1 −1 1

#9: Using LU-Factorization to Solve a System

Steps:

1. Solve Ly = b using forward substitution.

2. Solve Ux = y using back substitution.

Example:

Solve Ax = b for:
2 1 1 5
A = 4 −6 0 , ​ ​ ​ ​ ​
b = −2 . ​ ​ ​

−2 7 2 9

1. Use LU-factorization:

1 0 0 2 1 1
L= 2 1 0 ,
​ ​ ​ ​ ​ U = 0 −8 −2 . ​ ​ ​ ​ ​

−1 −1 1 0 0 1
2. Solve Ly = b:
1 0 0 y1 5 ​

2
​ 1 0 ​ ​ ​ ​ y2 = −2 .
​ ​ ​ ​ ​ ​ ​

−1 −1 1 y3 9 ​

5
Forward substitution yields y = −12 . ​ ​ ​

−2
3. Solve Ux = y:
2 1 1 x1 5 ​

0 −8 −2
​ ​ ​ ​ ​ x2 = −12 .
​ ​ ​ ​ ​ ​ ​

0 0 1 x3 −2 ​

3
Back substitution yields x = 2 .​ ​ ​

−2