MATRIX ALGEBRA
1. WHAT IS A MATRIX?
A matrix is defined as rectangular array of numbers, parameters, or variables. Matrix algebra provides a method to
solve a system of simultaneous equations:
1. It provides a compact way of writing an equation system.
2. It provides a procedure to test for the existence of a solution.
3. If the solution exists, it provides a method of finding that solution.
General format of a system of 𝑚 equations with 𝑛 variables:
𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑛 𝑥𝑛 = 𝑑1
𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑛 𝑥𝑛 = 𝑑2
∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙
𝑎𝑚1 𝑥1 + 𝑎𝑚2 𝑥2 + ⋯ + 𝑎𝑚𝑛 𝑥𝑛 = 𝑑𝑚
The matrix format of writing a system of equations is as follows.
𝑎11
𝑎21
𝐴=[ ⋮
𝑎𝑚1
𝑎12
𝑎22
⋮
𝑎𝑚2
⋯
⋯
⋮
⋯
𝑎1𝑛
𝑎2𝑛
⋯ ]
𝑎𝑚𝑛
𝑥1
𝑥2
𝑥=[ ⋮]
𝑥𝑛
𝑑1
𝑑
𝑑 = [ 2]
⋮
𝑑𝑚
The shorthand way of representing matrix 𝐴 is: 𝐴 = [𝑎𝑖𝑗 ]
Example 1:
6𝑥1 + 3𝑥2 + 1𝑥3 = 22
6𝑥1 + 4𝑥2 − 2𝑥3 = 12
4𝑥1 − 3𝑥2 + 5𝑥3 = 10
In matrix format the equation is presented as:
6
𝐴 = [1
4
3
4
−1
1
−2]
5
𝑥1
𝑥 = [ 𝑥2 ]
𝑥𝑛
22
𝑑 = [12]
10
The number of rows and number of columns of a matrix define the dimensions of the matrix. The dimension is shown
as 𝑚 × 𝑛. A matrix with 4 rows and 3 columns is a "4 by 3" matrix: 4 × 3.
A matrix with m rows and 1 column is called a column vector: 𝑚 × 1. Similarly, a matrix with 1 row and n columns is
called a row vector: 1 × 𝑛.
A system of simultaneous equations can be represented as: 𝑨𝒙 = 𝒅
The shorthand representation of the system of equations implies that the left-hand side of the equation is obtained as
the product of the coefficient matrix 𝐴 times the variable matrix 𝑥. The product of these two matrices produces a new
matrix. This new matrix, in turn, is equal to the constant matrix 𝑑.
To explain the product of two matrices and what makes two matrices equal we need to understand matrix operations.
LN1—Matrix Algebra
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2. MATRIX OPERATIONS
2.1. Equality of Matrices
Two matrices 𝐴 = [𝑎𝑖𝑗 ] and 𝐵 = [𝑏𝑖𝑗 ] are said to be equal if and only if 𝑎𝑖𝑗 = 𝑏𝑖𝑗 . For example,
𝐴=[
4
2
3
]
0
𝐵=[
4
2
3
]
0
𝐴=𝐵
2.2. Addition and Subtraction of Matrices
Two matrices can be added (subtracted) if and only if they are conformable for addition (subtraction). That is, they
have the same dimensions.
𝑎11
𝑎21
𝐴=[ ⋮
𝑎𝑚1
𝑎12
𝑎22
⋮
𝑎𝑚2
𝑎11 ± 𝑏11
𝑎21 ± 𝑏21
𝐴=[
⋮
𝑎𝑚1 ± 𝑏𝑚1
⋯
⋯
⋮
⋯
𝑎1𝑛
𝑎2𝑛
⋯ ]
𝑎𝑚𝑛
𝑎12 ± 𝑏12
𝑎22 ± 𝑏22
⋮
𝑎𝑚2 ± 𝑏𝑚2
𝑏11
𝑏21
𝐵=[
⋮
𝑏𝑚1
⋯
⋯
⋮
⋯
𝑏12
𝑏22
⋮
𝑏𝑚2
⋯
⋯
⋮
⋯
𝑏1𝑛
𝑏2𝑛
]
⋯
𝑏𝑚𝑛
𝑎1𝑛 ± 𝑏1𝑛
𝑎2𝑛 ± 𝑏2𝑛
]
⋯
𝑎𝑚𝑛 ± 𝑏𝑚𝑛
Example 2
4
𝐴 = [2
6
8
3
1 −2]
5
7
6
𝐵=[ 3
−6
10
2
7
10
𝐴+𝐵 =[ 5
0
5
4]
8
18
3
18
8
2]
15
(See the Excel file E375 CH4 Outline for the numeric example)
2.2.1. Scalar Multiplication
a scalar is a number. The scalar multiplication of a matrix involves multiplying each element of that matrix by that
number.
Example 3
𝑠=2
4
𝐴 = [2
6
8
1
5
3
−2]
7
8
𝑠𝐴 = [ 4
12
16
2
10
6
−4]
14
2.2.2. Multiplication of Matrices
To multiply two matrices 𝐴 and 𝐵, the matrices must by conformable for multiplication. In multiplying matrix 𝐴 by
matrix 𝐵, 𝐴 is the lead matrix and 𝐵 is the 𝑙𝑎𝑔 𝑚𝑎𝑡𝑟𝑖𝑥. 𝐴 and 𝐵 are conformable for multiplication if the number of
columns (column dimension) of the lead matrix 𝐴 is equal to the number of rows (row dimension) of the lag matrix 𝐵.
Consider the matrix A with dimensions (𝑚 × 𝑛) and matrix B with dimensions (𝑛 × 𝑝). A and 𝐵 are conformable for
multiplication because the column dimension of 𝐴 is 𝑛 and row dimension of 𝐵 is also 𝑛. The product matrix is 𝐴𝐵 =
𝐶 has dimensions (𝑚 × 𝑝).
A  B  C
( mn ) ( n p )
( m p )
Note that if 𝐵 were the lead matrix and 𝐴 the lag matrix, then 𝐵 and 𝐴 would not be conformable for multiplication.
Thus, the commutative property of multiplication of numbers does not apply to matrices.
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2.2.2.1.
How to multiply two matrices
Take two matrices A and B , which are conformable for multiplication. The product matrix is, then
( 3 2 )
( 2 2 )
A B  C
(32) ( 23)
(33)
𝑎11
𝑎
𝐴 = [ 21
𝑎31
𝑎12
𝑎22 ]
𝑎32
𝐵=[
𝑏11
𝑏21
𝑏12
𝑏22
𝑏13
]
𝑏23
To determine C you must perform the following operation:
𝑎11 𝑏11 + 𝑎12 𝑏21
𝐴𝐵 = 𝐶 = [𝑎21 𝑏11 + 𝑎22 𝑏21
𝑎31 𝑏11 + 𝑎32 𝑏21
𝑎11 𝑏12 + 𝑎12 𝑏22
𝑎21 𝑏12 + 𝑎12 𝑏22
𝑎31 𝑏12 + 𝑎12 𝑏22
Generally, we can express two matrices
𝑎11 𝑏13 + 𝑎12 𝑏23
𝑎21 𝑏13 + 𝑎22 𝑏23 ]
𝑎31 𝑏13 + 𝑎32 𝑏23
A and B conformable for multiplication as:
( mn )
( n p )
𝐴 = [𝑎𝑖𝑘 ] and 𝐵 = [𝑏𝑘𝑗 ]
where
𝑖 = 1,2, ⋯ 𝑚
𝑗 = 1,2, ⋯ 𝑝
𝑘 = 1,2, ⋯ , 𝑛
Then the product matrix is
𝑛
𝐶 = [𝐶𝑖𝑗 ] = [∑ 𝑎𝑖𝑘 𝑏𝑘𝑗 ]
𝑘=1
In the above specific case,
𝑖 = 1,2,3
𝑗 = 1,2,3
𝑘 = 1,2
For example, the element in the second row and third column of the product matrix is:
2
𝐶23 = ∑ 𝑎2𝑘 𝑏𝑘3 = 𝑎21 𝑏13 + 𝑎22 𝑏23
𝑘=1
Example 4
4
𝐴 = [6
2
9
3]
5
𝐵=[
8
4
4×8+9×4
𝐶 = 𝐴𝐵 = [6 × 8 + 3 × 4
2×8+5×4
1
7
3
]
6
4×1+9×7
6×1+3×7
2×1+5×7
4×3+9×6
68
6 × 3 + 3 × 6] = [60
2×3+5×6
36
67
27
37
66
36]
36
In Excel to find the product 𝐴𝐵 use the =𝑀𝑀𝑈𝐿𝑇 command. =𝑀𝑀𝑈𝐿𝑇 is an “array” formula, requiring the following
steps:
1.
2.
3.
Highlight a 3-by-3 block of cells.
Call up =MMULT(array1, array2). “array 1” is the block of cells containing the elements of 𝐴 and “array2” is
the block containing elements of 𝐵.
Hold Ctrl and Shift keys and press Enter.
LN1—Matrix Algebra
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Example 5
0
6
𝐴=[
9
5
1
1
3
2
3
7
]
8
4
7
𝐵 = [5
7
4
8]
6
A B  C
( 43) (32)
( 42)
26
96
𝐶=[
134
73
26
74
]
108
60
2.3. Back to the simultaneous equation system
Recall the system of equations,
6𝑥1 + 3𝑥2 + 1𝑥3 = 22
6𝑥1 + 4𝑥2 − 2𝑥3 = 12
4𝑥1 − 3𝑥2 + 5𝑥3 = 10
In matrix format the equation is presented as:
6
3
A = [1
4
(33)
4 −1
𝑥1
𝑥
x = [ 2]
( 31)
𝑥𝑛
1
−2]
5
22
d = [12]
( 31)
10
Using the matrix multiplication procedure, we have
6
A x = [1
(33) (31)
4
3
4
−1
6𝑥1 + 3𝑥2 + 2𝑥3
1 𝑥1
−2] [ 𝑥2 ] = [6𝑥1 + 4𝑥2 − 2𝑥3 ]
4𝑥1 − 4𝑥2 + 5𝑥3
5 𝑥𝑛
The product matrix 𝐴𝑥 has the dimensions (3 × 1) and is equal to the constant matrix d .
( 31)
6𝑥1 + 3𝑥2 + 2𝑥3
22
[6𝑥1 + 4𝑥2 − 2𝑥3 ] = [12]
4𝑥1 − 4𝑥2 + 5𝑥3
10
3. IDENTITY MATRICES
An identity matrix is a square matrix with 1s in its principal diagonal and 0s everywhere else.
𝐼2 = [
1
0
0
]
1
1
𝐼3 = [0
0
0
1
0
0
0]
1
The identity matrix plays a role similar to 1 in scalar algebra (regular number system). You can pre or post multiply a
matrix 𝐴 by an identity matrix and obtain the same result—the original matrix A.
𝐼𝐴 = 𝐴𝐼 = 𝐴
Example 6
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8
A =[
9
23
3
5
1
I =[
0
0
]
1
22 
0
]
7
1 0
I = [0 1
33
0 0
8
A =[
9
0
0]
1
3
5
23
8
AI = [
9
3
5
0
]
7
8
IA = [
9
3
5
0
]
7
23
0
]
7
23
4. TRANSPOSE OF A MATRIX
The transpose of matrix 𝐴, denoted by 𝐴′, is obtained by interchanging the rows with columns.
Example 7
The following are several examples of matrix transposes.
𝐴=[
8
9
3
5
0
]
7
8
𝐴′ = [3
0
9
5]
7
7
𝐵 = [5
7
4
8]
6
7
𝐵′ = [
4
5
8
7
]
6
6
𝐶 = [9
5
1 7
3 8]
2 4
6
𝐶 ′ = [1
7
9
3
8
5
2]
4
5. INVERSE OF A MATRIX
The inverse of a square matrix 𝐴, denoted by 𝐴−1 , is a matrix that satisfies the following condition:
𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼
Example 8
6
𝐴 = [9
5
1 7
3 8]
2 4
1 0
𝐴𝐴−1 = [0 1
0 0
0
0]
1
−1
𝐴−1 = [ 4
3
10
−11
−7
1
𝐴−1 𝐴 = [0
0
0
1
0
−13
15]
9
0
0]
1
To find the inverse of 𝐴 above I have used the Excel array function =𝑀𝐼𝑁𝑉𝐸𝑅𝑆𝐸. Finding the inverse of a matrix, if it
exists, without a computer is an involved process. The process is explained below.
If the inverse of a matrix A exists, then it is called a nonsingular matrix. If the inverse does not exist, then A is called a
singular matrix.
5.1. Properties of Inverse Matrices
1.
2.
3.
(𝐴−1 )−1 = 𝐴
The inverse of an inverse matrix is the original matrix:
The inverse of product of the lead matrix 𝐴 and the lag matrix 𝐵 is equal to the product of the inverse of the
(𝐴𝐵)−1 = 𝐵 −1 𝐴−1
lead matrix 𝐵 −1 and the inverse of lag matrix 𝐴−1 :
(𝐴′)−1 = (𝐴−1 )′
Inverse of the transpose is equal to the transpose of the inverse:
5.2. Inverse Matrix and Solution of Linear Equation System
LN1—Matrix Algebra
Page 5 of 12
It was shown that a linear equation system can be presented in the following matrix format:
x  d
A
( mm) ( m1)
( m1)
Now pre-multiply both sides by 𝐴−1 :
𝐴−1 𝐴𝑥 = 𝐴−1 𝑑
Given that 𝐴−1 𝐴 = 𝐼, then
𝐴−1 𝐴𝑥 = 𝐼𝑥 = 𝑥
Thus,
𝑥 = 𝐴−1 𝑑
The matrix product
A 1 d
( mm) ( m1)
yields an (𝑚 × 1) matrix whose elements are the solutions for the equation system.
Example 9
Using the equation system from Example 1 we have
6
[1
4
3
4
−1
1 𝑥1
22
−2] [ 𝑥2 ] = [12]
5 𝑥𝑛
10
Using the =𝑀𝐼𝑁𝑉𝐸𝑅𝑆𝐸 command in Excel we can find the 𝐴−1 .
0.3462
𝐴−1 = [−0.2500
−0.3269
−0.3077
0.5000
0.3462
−0.1923
0.2500]
0.4038
Thus, the solution matrix is obtained by finding the product of 𝐴−1 and d on the right-hand-side (using the =𝑀𝑀𝑈𝐿𝑇
command):
𝑥1
0.3462
[ 𝑥2 ] = [−0.2500
𝑥𝑛
−0.3269
−0.3077
0.5000
0.3462
−0.1923 22
2
0.2500] [12] = [3]
0.4038 10
1
Finding the inverse matrix is yet to be explained. This is what comes next.
5.3. Finding the Inverse Matrix
Clearly, for a system of linear equations to have a set of unique solutions, the coefficient matrix 𝐴 must have an
inverse. In other words, 𝐴 must be a nonsingular matrix.
5.3.1. The Requirements for Non-singularity of a Matrix
In order for matrix 𝐴 to be nonsingular all of its rows must be linearly independent. This means that none of the rows
can be a linear combination of other rows. Consider, for example the following equation system.
6𝑥1 + 13𝑥2 + 1𝑥3 = 22
6𝑥1 + 14𝑥2 − 2𝑥3 = 12
LN1—Matrix Algebra
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8𝑥1 + 11𝑥2 − 3𝑥3 = 10
The coefficient matrix is
6
[1
8
3
4
11
1
−2]
−3
Note that, although not obvious, the third row is a linear combination of the first two rows:
[8
−3] = [6 + 2 × 1
11
3+2×4
1 − 2 × 2]
This matrix does not have an inverse. If you try to find the inverse in Excel, you will receive an error message.
5.3.2. Using the Determinant of a Matrix to Test for Singularity
The determinant of a square Matrix, denoted by |𝐴| is a uniquely defined number or numeric value associated with
that matrix. Let’s start first with a (2 × 2) matrix and show how to find the determinant:
𝑎11
𝐴 = [𝑎
21
𝑎12
𝑎22 ]
𝑎
|𝐴| = |𝑎11
𝑎12
𝑎22 | = 𝑎11 𝑎22 − 𝑎12 𝑎21
21
Example 10
𝐴=[
10
4
|𝐴| = |
8
]
5
10 8
| = 10 × 5 − 8 × 4 = 18
4 5
The determinant of a (3 × 3) matrix A
𝑎11
𝑎
𝐴 = [ 21
𝑎31
𝑎12
𝑎22
𝑎32
𝑎13
𝑎23 ]
𝑎33
is determined as follows:
𝑎11
|𝐴| = |𝑎21
𝑎31
𝑎12
𝑎22
𝑎32
𝑎13
𝑎
𝑎23 | = 𝑎11 | 22
𝑎32
𝑎33
𝑎23
𝑎21
𝑎33 | − 𝑎12 |𝑎31
𝑎23
𝑎21
𝑎33 | + 𝑎13 |𝑎31
𝑎22
𝑎32 |
|𝐴| = 𝑎11 (𝑎22 𝑎33 − 𝑎23 𝑎32 ) − 𝑎12 (𝑎21 𝑎33 − 𝑎23 𝑎31 ) + 𝑎13 (𝑎21 𝑎32 − 𝑎22 𝑎31 )
Example 11
6
|𝐴| = |9
5
1 7
3
3 8| = 6 |
2
2 4
8
9
|−|
4
5
8
9
| + 7|
4
5
3
|
2
|𝐴| = 6(3 × 4 − 8 × 2) − (9 × 4 − 8 × 5) + 7(9 × 2 − 3 × 5) = 1
Note that the above “3rd order” determinant is expanded into an expression containing three “2 nd order”
𝑎22 𝑎23
𝑠𝑢𝑏determinants. For example, the subdeterminant |𝑎
𝑎33 | is obtained by deleting the third row and third column
32
LN1—Matrix Algebra
Page 7 of 12
of |𝐴|. This subdeterminant is called the minor of the element 𝑎11 , which is the element located in first row and first
column of |𝐴|. The minor of 𝑎11 is denoted by |𝑀11 |.
𝑎11
|𝑀11 | = |𝑎21
𝑎31
𝑎12
𝑎22
𝑎32
𝑎13
𝑎
𝑎23 | = | 22
𝑎32
𝑎33
𝑎23
𝑎33 |
In general, |𝑀𝑖𝑗 | represents the minor of the element 𝑎𝑖𝑗 , which is obtained by deleting the 𝑖th row and the 𝑗th column.
Thus, we can write the 3rd order determinant |𝐴| above as:
|𝐴| = 𝑎11 |𝑀11 | − 𝑎12 |𝑀12 | + 𝑎13 |𝑀13 |
A minor with an algebraic sign attached to it is called the cofactor of a given element 𝑎𝑖𝑗 . The “+” or “−“ sign depends
whether the sum 𝑖 + 𝑗 is even or odd:
|𝐶𝑖𝑗 | ≡ (−1)𝑖+𝑗 |𝑀𝑖𝑗 |
Thus, the determinant |𝐴|, obtained through the expansion by the first row can be represented as,
3
|𝐴| = ∑ 𝑎1𝑗 |𝐶1𝑗 |
𝑗=1
Using the cofactor notation the determinant is expressed as:
|𝐴| = 𝑎11 |𝐶11 | + 𝑎12 |𝐶12 | + 𝑎13 |𝐶13 |
You can expand any nth order determinant by any row or any column.
determinant |B| by the second row we have:
For example, expanding a 4th order
4
|𝐵| = ∑ 𝑏2𝑗 |𝐶2𝑗 |
𝑗=1
In general,
𝑛
|𝐴| = ∑ 𝑎𝑖𝑗 |𝐶𝑖𝑗 |
(expansion by 𝑖th row)
𝑗=1
𝑛
|𝐴| = ∑ 𝑎𝑖𝑗 |𝐶𝑖𝑗 |
(expansion by 𝑗th column)
𝑖=1
1.
5.3.3. Basic Properties of Determinants
The determinant of the transpose A′ is the same as the determinant of A
|𝐴| = |5
4
2.
3
| = 28
8
|𝐴′| = |5
3
4
| = 28
8
Interchange of any two rows (or any two) columns will change the algebraic sign of the determinant,
|𝐴| = |5
4
LN1—Matrix Algebra
3
| = 28
8
|𝐵| = |
4
5
8
| = −28
3
Page 8 of 12
3.
Multiplication of any one row (or one column) by a scalar k will change the value of the determinant k-fold.
|𝐴| = |5
4
4.
3
| = 28
8
2×3
| = 2 × 28 = 56
8
The addition of a multiple of any row (column) to another row (column) will leave the value of the
determinant unchanged.
|𝐴| = |5
4
5.
|𝐵| = |2 × 5
4
3
| = 28
8
|𝐵| = |
5
4+2×5
3
| = 5 × 14 − 3 × 14 = 28
8+2×3
If one row (or column) is a multiple of another row (or column) the determinant vanishes—it is zero.
|𝐴| = | 5
2×5
3
| = 2(5 × 3 − 5 × 3) = 0
2×3
By extension, if one row (column) is a linear combination of another row (column) or a linear combination
any two rows (columns) the determinant vanishes.
|𝐴| = |
5
5+5×𝑘
6
|𝐵| = | 1
6+𝑘
|𝐵| = 6 |
3
| = 15 + 15𝑘 − 15 − 15𝑘 = 0
3+3×𝑘
3
4
3 + 4𝑘
4
3 + 4𝑘
1
−2 | = 6|𝐶11 | + 3|𝐶12 | + |𝐶13 |
1 − 2𝑘
−2
1
| + 3(−1) |
1 − 2𝑘
6+𝑘
−2
1
|+|
1 − 2𝑘
6+𝑘
4
|
3 + 4𝑘
|𝐵| = 6(4 − 8𝑘 + 6 + 8𝑘) − 3(1 − 2𝑘 + 12 + 2𝑘) + (3 + 4𝑘 − 24 − 4𝑘)
|𝐵| = 6(10) − 3(13) + (−21) = 0
6.
The expansion of a determinant by alien cofactors yields a value of zero.
6
|𝐴| = |9
5
1
3
2
7
8|
4
Multiply the elements of the first row by the cofactors of the elements of second row.
3
∑ 𝑎1𝑗 |𝐶2𝑗 | = 𝑎11 |𝐶21 | + 𝑎12 |𝐶22 | + 𝑎13 |𝐶23 |
𝑗=1
3
∑ 𝑎1𝑗 |𝐶2𝑗 | = 6(−1) |
𝑗=1
LN1—Matrix Algebra
1
2
7
6
|+|
4
5
7
6
| + 7(−1) |
4
5
1
| = −6(−10) − (11) − 7(7) = 0
2
Page 9 of 12
5.3.4. Criterion for Non-singularity of a Matrix—Non-vanishing Determinant
As mentioned in Section 5.3.1.: In order for matrix 𝐴 to be nonsingular all of its rows must be linearly independent.
This means that none of the rows can be a linear combination of other rows. There you have it! If any row of the
determinant of the coefficient matrix of system of linear equations is a linear combination of any two rows, or a
multiple of any other row, the determinant vanishes, the matrix is singular and the inverse of the matrix does not
exist.
5.4. How to Find the Inverse of Matrix A
Let’s start with a (3 × 3) matrix:
𝑎11
𝐴 = [𝑎21
𝑎31
1.
𝑎12
𝑎22
𝑎32
𝑎13
𝑎23 ]
𝑎33
Form the cofactor matrix by replacing each element by its cofactor.
|𝐶11 |
𝐶 = [|𝐶21 |
|𝐶31 |
2.
|𝐶12 |
|𝐶22 |
|𝐶32 |
|𝐶13 |
|𝐶23 |]
|𝐶33 |
Form the transpose of the cofactor matrix: 𝐶′. This called the adjoint of matrix A and is donate by adj 𝐴.
|𝐶11 |
𝐶 = adj 𝐴 = [|𝐶12 |
|𝐶13 |
′
3.
|𝐶21 |
|𝐶22 |
|𝐶23 |
|𝐶31 |
|𝐶32 |]
|𝐶33 |
Pre multiply adj 𝐴 by matrix 𝐴
𝑎11
𝐴𝐶′ = [𝑎21
𝑎31
𝑎12
𝑎22
𝑎32
𝑎13 |𝐶11 |
𝑎23 ] [|𝐶12 |
𝑎33 |𝐶13 |
3
|𝐶21 |
|𝐶22 |
|𝐶23 |
3
|𝐶31 |
|𝐶32 |]
|𝐶33 |
3
∑ 𝑎1𝑗 |𝐶1𝑗 |
∑ 𝑎1𝑗 |𝐶2𝑗 |
∑ 𝑎1𝑗 |𝐶3𝑗 |
𝑗=1
3
𝑗=1
3
𝑗=1
3
∑ 𝑎2𝑗 |𝐶2𝑗 |
∑ 𝑎2𝑗 |𝐶3𝑗 |
𝑗=1
3
𝑗=1
3
∑ 𝑎3𝑗 |𝐶2𝑗 |
∑ 𝑎3𝑗 |𝐶3𝑗 |
]
𝑗=1
𝐴𝐶′ = ∑ 𝑎2𝑗 |𝐶1𝑗 |
𝑗=1
3
∑ 𝑎3𝑗 |𝐶1𝑗 |
[𝑗=1
𝑗=1
Note that in the three elements of the principal diagonal of AC′ all are determinants of matrix 𝐴, and the other
elements are expansions by alien cofactors and thus are all equal to zero.
|𝐴| 0
𝐴𝐶′ = [ 0 |𝐴|
0
0
0
0]
|𝐴|
Now you can view 𝐴𝐶′ as the identity matrix multiplied by the scalar |𝐴|.
1
𝐴𝐶 ′ = |𝐴| [0
0
LN1—Matrix Algebra
0
1
0
0
0] = |𝐴|𝐼
1
Page 10 of 12
𝐴𝐶 ′ = |𝐴|𝐼
4.
Multiply both sides by the scalar 1⁄|𝐴|.
𝐴𝐶′
=𝐼
|𝐴|
or
𝐴
𝐶′
=𝐼
|𝐴|
Then pre multiply both sides by 𝐴−1 :
𝐴−1 𝐴
𝐶′
= 𝐴−1 𝐼
|𝐴|
𝐶′
= 𝐴−1
|𝐴|
Now we have obtained the inverse of 𝐴
𝐴−1 =
𝐶′
1
=
adj 𝐴
|𝐴| |𝐴|
Example 12
Find the inverse of the coefficient matrix from Example 1. Show each step.
6
𝐴 = [1
4
3
4
−1
1
−2]
5
The cofactor matrix is:
4 −2
|
−1
5
3 1
𝐶 = −|
|
−1 5
3
1
[ |4 −2|
−2
|
5
6 1
|
|
4 5
6
1
−|
|
1 −2
|
−|
1
4
1
4
6
−|
4
|
4
|
−1
18
3
| = [−16
−1
−10
6 3
|
|]
1 4
−13
26
13
−17
18]
21
Adjoint of A is:
18
𝐶 ′ = [−13
−17
−16
26
18
−10
13]
21
Now pre multiply 𝐶′ by 𝐴.
6
3
𝐴𝐶′ = [1
4
4 −1
1
18
−2] [−13
5 −17
52 0
𝐴𝐶′ = [ 0 52
0
0
−16
26
18
−10
13]
21
0
0]
52
The determinant of A is the element in the principal diagonal of 𝐴𝐶′.
|𝐴| = 52
LN1—Matrix Algebra
Page 11 of 12
Thus,
𝐴−1 =
18
𝐶′
1
= ( ) [−13
|𝐴|
52
−17
0.3462
𝐴−1 = [−0.2500
−0.3269
−16
26
18
−0.3077
0.5000
0.3462
−10
13]
21
−0.1923
0.2500]
0.4038
You may check the result by applying the =MINVERSE command in Excel.
.
LN1—Matrix Algebra
Page 12 of 12