The Inverse of a Square Matrix

Tutorial on how to find the inverse of a square matrix.

Definition of The Matrix Identity

The identity matrix I n is the square matrix with order n x n and with the elements in the main diagonal consiting of 1's and all other elements are equal to zero.

Examples:

The 2 x 2 identity matrix

``` I 2 =  [1    0]

[0    1]

```

The 3 x 3 identity matrix

```  [1     0     0]

[0     1     0]

[0     0     1]

```

In general the n x n matrix identity has the form

```       [1     0    ...    0]

[0     1    ...    0]

[0     0    ...    0]

A  =   [.                  ]

[.                  ]

[0     0     ...   1]

```

Definition of The Inverse of a Matrix

Let A be a square matrix of orde n x n. If there exists a matrix B such that

A B = I
n = B A

Then B is called the inverse matrix of A. (and of course matrix A is the inverse matrix of B).

Examples:

Verify that matrices A and B given below are inverses of each other.

``` A = [2     3]

[3     4]

```

``` B = [-4    3]

[3    -2]

```

Solution

Let us find the products AB and BA

```AB = [2     3][-4    3] = [-8 + 9     6 - 6] = [1   0]

[3     4][3    -2]   [-12 + 12   9 - 8]   [0   1]

```

```BA = [-4     3][2    3] = [-8 + 9   -12 + 12] = [1   0]

[3     -2][3    4]   [6 - 6       9 - 8]   [0   1]

```

We have verified that AB = BA = I
2 and therefore A and B are inverse of each other.

The inverse of a square matrix A is denoted as A
-1.

How to Find The Inverse of a 2 by 2 Square Matrix?

Example 1: Find a formula for the inverse of a 2 x 2 square matrix A given by

```A = [a   b]

[c   d]

```

Solution

Let B be the inverse of A given by

```B = [a'   b']

[c'   d']

```

From definition, we have

AB = I

```[a   b][a'   b'] = [1   0]

[c   d][c'   d']   [0   1]

```

We need to find terms a', b', c' and d' of B (inverse) in terms of a, b, c and d which are the terms of matrix A. Multiply matrices A and B above.

```[a a' + b c'    a b' + b d']= [1   0]

[c a' + d c'    c b' + d d']  [0   1]

```

The above gives 4 equations

a a' + b c' = 1 (equation 1)
c a' + d c' = 0 (equation 2)
a b' + b d' = 0 (equation 3)
c b' + d d' = 1 (equation 4)

We now need to solve the first and second equations simultaneously to find a' and c' in terms of a, b, c and d.

Multiply the first equation by d and the second equation by b and subtract the left and right terms of the equations obtained to find

a'(a d - b c) = d

a ' = d / (a d - b c)

Substitute a' in the second equation to obtain c'= - c / (a d - b c)

To find b' and d', multiply equation (3) by d and equation (4) by b and subtract the left and right terms of the equations obtained to find

b'(a d - b c) = - b

or b' = - b / (a d - b c)

Substitute b' in equation (3) to obtain d'= a / (a d - b c)

Finally matrix B , the inverse of A, is given by

```B = 1 / (a d - b c) [d   -b]

[-c   a]

```
In solving the question in the above example, we have found a formula for the inverse of any invertible 2 x 2 matrix. A 2 x 2 matrix will have an inverse if it determinant D = a d - b c is not equal to zero since division by zero is not allowed.

Example 2: Use the above formula for the inverse of a 2 x 2 square matrix to find the inverse of

```A = [3   5]

[2   3]

```

Solution

Acoording to formula above the inverse of A , A
-1 is given by

```A -1 = 1 / [ (3)(3) - (5)(2) ] [ 3    -5 ]

[-2      3]

```
Simplify to obtain
```A -1 =  [ -3    5 ]

[2      -3]

```
As an exercise, show that the product A A -1 = A -1 A = I 2.

How to Find The Inverse of a 3 by 3 Square Matrix?

Example 1: Find the inverse of a 3 x 3 square matrix A given by

```    [1   1   2]

A = [2   4  -3]

[3   6  -5]

```

Solution

We use a method based on augmenting the given matrix by the identity matrix as follows

```[1   1   2   |   1    0    0]    row(1)

[2   4  -3   |   0    1    0]    row(2)

[3   6  -5   |   0    0    1]    row(3)

```

We next use basic row operations so that the identity matrix on the right is moved to the left

We multiply row(1) by - 2 and add it to row(2) and put the result back in row(2). We also multiply row(1) by - 3 and add it to row(3) and put the result back in row(3).

```[1   1    2   |   1    0    0]    row(1)

[0   2   -7   |   -2   1    0]    row(2)

[0   3   -11  |   -3   0    1]    row(3)

```
We multiply row(2) by 3 and row(3) by -2, add them and put the row obtained in row(3).

```[1   1    2   |   1    0    0]    row(1)

[0   2   -7   |   -2   1    0]    row(2)

[0   0    1   |    0   3   -2]    row(3)

```
We multiply row(1) by -2 and add it to row(2) and put the row obtained in row(1).
```[-2  0  -11   |   -4   1    0]    row(1)

[0   2   -7   |   -2   1    0]    row(2)

[0   0    1   |    0   3   -2]    row(3)

```
We multiply row(3) by 11 and add it to row(1) and put the row obtained in row(1).
```[-2  0    0   |   -4   34  -22]    row(1)

[0   2   -7   |   -2   1     0]    row(2)

[0   0    1   |    0   3    -2]    row(3)

```
We multiply row(3) by 7 and add it to row(2) and put the row obtained in row(2).
```[-2  0    0   |   -4   34  -22]    row(1)

[0   2    0   |   -2   22  -14]    row(2)

[0   0    1   |    0   3    -2]    row(3)

```
We now multiply all terms in row(1) by -1/2 and multiply all terms in row(2) by 1/2.
```[1  0    0    |   2   -17  11]    row(1)

[0   1    0   |   -1   11  -7]    row(2)

[0   0    1   |    0   3   -2]    row(3)

```
Now that we have the identity matrix on the left, the 3 by 3 matrix on the right is the inverse matrix. The inverse of A is given by
```               [2   -17  11]

A -1 = [-1   11  -7]

[0    3   -2]

```
A similar method based on row operations can in principle be used to find the inverse of any invertible (that has an inverse) square matrix.

Exercises

Find the inverse of matrices A and B given below.

```A =  [2    3]

[3    5]

```

```     [1   1    2]

B =  [1   2    4]

[2  -1    0]

```

```A -1 =  [5   -3]

[-3    2]

```

```       [2    -1      0]

B -1 =  [4    -2     -1]

[-5/2  3/2   1/2]

```

More pages and references related to matrices.