Math 15 (Notes) encoding

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 =	é ê ë		3 1 0		-1 1 2		0 2 4		ù ú û

Solution:

é ê ë 3 1 0 -1 1 2 0 2 4 ï ï ï 1 0 0 0 1 0 0 0 1 ù ú û		write the augmented matrix [ A | I ]
é ê ë 1 2 0 1 -1 2 2 0 4 ï ï ï 0 1 0 1 0 0 0 0 1 ù ú û		Swap rows 1 and 2 (to bring a 1 to the upper left)
é ê ë 1 0 0 1 -4 2 2 -6 4 ï ï ï 0 1 0 1 -3 0 0 0 1 ù ú û		Add -3 times row 1 to row 2 (to get a 0 in the first column)
é ê ë 1 0 0 1 2 -4 2 4 -6 ï ï ï 0 0 1 1 0 -3 0 1 0 ù ú û		Swap second and third rows (this will make the computation easier)
é ê ë 1 0 0 1 2 0 2 4 2 ï ï ï 0 0 1 1 0 -3 0 1 2 ù ú û		Add 2 times row 2 to row 3 (to get a zero in the second column)
é ê ë 1 0 0 1 1 0 2 2 1 ï ï ï 0 0 1/2 1 0 -3/2 0 1/2 1 ù ú û		Divide rows 2 and 3 by 2 (to get leading ones)
é ê ë 1 0 0 0 1 0 0 2 1 ï ï ï 0 0 1/2 1 0 -3/2 -1/2 1/2 1 ù ú û		Add -1 times row 2 to row 1 (to get zeros above the leading 1's)
é ê ë 1 0 0 0 1 0 0 0 1 ï ï ï 0 -1 1/2 1 3 -3/2 -1/2 -3/2 1 ù ú û		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 =	é ê ë		0 -1 1/2		1 3 -3/2		-1/2 -3/2 1		ù ú û

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		=		é ê ë 0 -1 1/2 1 3 -3/2 -1/2 -3/2 1 ù ú û é ê ë 0 -1 1/2 1 3 -3/2 -1/2 -3/2 1 ù ú û
			=		é ê ë 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) ù ú û
			=		é ê ë 0+1+0 -3+3+0 3/2-3/2+0 0+1-1 1+3-3 -(1/2)-(3/2)+1 0+2-2 0+6-6 0-3+4 ù ú û
			=		é ê ë 1 0 0 0 1 0 0 0 1 ù ú û
			=		I

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

Math 15 (Spring 2003) web pages Created: 28 Mar 2003 Last modified: Jun 6, 2003 9:38:36 AM Comments to: dpvc@union.edu

Selected Course Notes Gauss-Jordan Elimination Notes