Math 15 (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 matrix 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

 A = éêë 310 -112 024 ùúû

## Solution:

 éêë 310 -112 024 ïïï 100 010 001 ùúû
write the augmented matrix [ A | I ]
 éêë 120 1-12 204 ïïï 010 100 001 ùúû
Swap rows 1 and 2
(to bring a 1 to the upper left)
 éêë 100 1-42 2-64 ïïï 010 1-30 001 ùúû
Add -3 times row 1 to row 2
(to get a 0 in the first column)
 éêë 100 12-4 24-6 ïïï 001 10-3 010 ùúû
Swap second and third rows
(this will make the computation easier)
 éêë 100 120 242 ïïï 001 10-3 012 ùúû
Add 2 times row 2 to row 3
(to get a zero in the second column)
 éêë 100 110 221 ïïï 001/2 10-3/2 01/21 ùúû
Divide rows 2 and 3 by 2
 éêë 100 010 021 ïïï 001/2 10-3/2 -1/21/21 ùúû
Add -1 times row 2 to row 1
(to get zeros above the leading 1's)
 éêë 100 010 001 ïïï 0-11/2 13-3/2 -1/2-3/21 ùúû
Add -2 times row 3 to row 2
(to get a zero above the leading 1)

This has reduced the left-hand side to the identity matrix, so the right-hand side is the inverse of A.

 A-1 = éêë 0-11/2 13-3/2 -1/2-3/21 ùúû

To check the answer, we compute the product A-1A and see if it equals I. If it does, then we have the inverse.

A-1A=
 éêë 0-11/2 13-3/2 -1/2-3/21 ùúû éêë 0-11/2 13-3/2 -1/2-3/21 ùúû
=
 éêë 0(3)+1(1)+(-1/2)(0)-1(3)+3(1)+(-3/2)(0)(1/2)(3)+(-3/2)(1)+1(0) 0(-1)+1(1)+(-1/2)(2)-1(-1)+3(1)+(-3/2)(2)(1/2)(-1)+(-3/2)(1)+1(2) 0(0)+1(2)+(-1/2)(4)-1(0)+3(2)+(-3/2)(4)(1/2)(0)+(-3/2)(2)+1(4) ùúû
=
 éêë 0+1+0-3+3+03/2-3/2+0 0+1-11+3-3-(1/2)-(3/2)+1 0+2-20+6-60-3+4 ùúû
=
 éêë 100 010 001 ùúû
=I

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

 Math 15 (Winter 2001) web pages Created: 09 Feb 1998 Last modified: 11 Mar 2001 23:59:59 Comments to: `dpvc@union.edu`