**Up:** *Math 15 Selected Class Notes*

# Matrix Inverse Notes:

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

- 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.

- 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*^{-1}*A*
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*^{-1}*A* = I, we know we have the
correct inverse.

**Up:** *Math 15 Selected Class Notes*

Comments to:
dpvc@union.edu

Created: Feb 9 1998 ---
Last modified: Feb 9, 1998 10:23:21 AM