Math 15H (Notes)

# Matrix Inverse Notes:

The procedure for finding an inverse of a square matrix, A, is as follows:
1. Form the augmented matrix that has A on the left and the identity matrix, I, on the right.

2. Perform Gauss-Jordan elimination on the augmented matrix. The reduced row echelon form will either be an identity matric on the left, or will have a row of zeros on the bottom of the left half of the augmented matrix.

3. If the procedure produces a row of zeros, then the matrix A is not invertible. If it produces an itendity matrix, then the matrix on the right of the augmented matrix is the inverse, A-1.

## Example:

Find the inverse of the matrix
```        +-         -+
|  3 -1  0  |
A = |  1  1  2  |
|  0  2  4  |
+-         -+

```
Solution:
```
+-                     -+
|  3 -1  0  |  1  0  0  |     write down the augmented matrix
|  1  1  2  |  0  1  0  |      [ A | I ]
|  0  2  4  |  0  0  1  |
+-                     -+

+-                     -+
|  1  1  2  |  0  1  0  |     Swap rows 1 and 2
|  3 -1  0  |  1  0  0  |     (to bring a 1 to the upper left)
|  0  2  4  |  0  0  1  |
+-                     -+

+-                     -+
|  1  1  2  |  0  1  0  |     Add -3 times row 1 to row 2
|  0 -4 -6  |  1 -3  0  |     (to get a 0 in the first column)
|  0  2  4  |  0  0  1  |
+-                     -+

+-                     -+
|  1  1  2  |  0  1  0  |     Swap second and third rows
|  0  2  4  |  0  0  1  |     (this will make the computation easier)
|  0 -4 -6  |  1 -3  0  |
+-                     -+

+-                       -+
|  1  1  2  |  0  1  0    |   Divide second row by 2
|  0  1  2  |  0  0  1/2  |   (to get a leading 1)
|  0 -4 -6  |  1 -3  0    |
+-                       -+

+-                       -+
|  1  0  0  |  0  1 -1/2  |   Add -1 times row 2 to row 1
|  0  1  2  |  0  0  1/2  |   Add 4 times row 2 to row 3
|  0  0  2  |  1 -3  2    |   (to get zeros in column 2)
+-                       -+

+-                        -+
|  1  0  0  |  0   1  -1/2 |  Divide third row by 2
|  0  1  2  |  0   0   1/2 |  (to get a leading 1)
|  0  0  1  | 1/2 -3/2  1  |
+-                        -+

+-                        -+
|  1  0  0  |  0   1  -1/2 |  Add -2 times row 3 to row 2
|  0  1  0  | -1   3  -3/2 |  (to get a zero in column 3)
|  0  0  1  | 1/2 -3/2  1  |
+-                        -+

```
This has reduced the left-hand side to the identity matrix, so the right-hand side is the inverse of A.
```
+-               -+
-1    |  0    1   -1/2  |
A    =  | -1    3   -3/2  |
| 1/2  -3/2   1   |
+-               -+
```
To check the answer, we compute the product A-1A and see if it equals I. If it does, then we have the inverse.
```
+-              -+  +-         -+
-1     |  0    1  -1/2  |  |  3 -1  0  |
A    A = | -1    3  -3/2  |  |  1  1  2  |
| 1/2 -3/2  1/2  |  |  0  2  4  |
+-              -+  +-         -+

+-                                                                         -+
|    0(3)+1(1)+(-1/2)(0)     0(-1)+1(1)+(-1/2)(2)     0(0)+1(2)+(-1/2)(4)   |
= |   -1(3)+3(1)+(-3/2)(0)    -1(-1)+3(1)+(-3/2)(2)    -1(0)+3(2)+(-3/2)(4)   |
|  (1/2)(3)+(-3/2)(1)+1(0) (1/2)(-1)+(-3/2)(1)+1(2) (1/2)(0)+(-3/2)(2)+1(4) |
+-                                                                         -+

+-                                  -+
|    0+1+0        0+1-1       0+2-2  |
= |   -3+3+0        1+3-3       0+6-6  |
|  3/2-3/2+0  -(1/2)-(3/2)+1  0-3+4  |
+-                                  -+

+-          -+
|   1  0  0  |
= |   0  1  0  |
|   0  0  1  |
+-          -+

= I

```
Since A-1A = I, we know we have the correct inverse.

 Math 15H (Fall 2000) web pages Created: 09 Feb 1998 Last modified: 10 Nov 2000 11:06:13 Comments to: `dpvc@union.edu`