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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 =

3
1
0
-1
1
2
0
2
4


Solution:



3
1
0
-1
1
2
0
2
4


1
0
0
0
1
0
0
0
1


write the augmented matrix [ A | I ]


1
2
0
1
-1
2
2
0
4


0
1
0
1
0
0
0
0
1


Swap rows 1 and 2
(to bring a 1 to the upper left)


1
0
0
1
-4
2
2
-6
4


0
1
0
1
-3
0
0
0
1


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


1
0
0
1
2
-4
2
4
-6


0
0
1
1
0
-3
0
1
0


Swap second and third rows
(this will make the computation easier)


1
0
0
1
2
0
2
4
2


0
0
1
1
0
-3
0
1
2


Add 2 times row 2 to row 3
(to get a zero in the second column)


1
0
0
1
1
0
2
2
1


0
0
1/2
1
0
-3/2
0
1/2
1


Divide rows 2 and 3 by 2
(to get leading ones)


1
0
0
0
1
0
0
2
1


0
0
1/2
1
0
-3/2
-1/2
1/2
1


Add -1 times row 2 to row 1
(to get zeros above the leading 1's)


1
0
0
0
1
0
0
0
1


0
-1
1/2
1
3
-3/2
-1/2
-3/2
1


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 =

0
-1
1/2
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.

A-1A=


0
-1
1/2
1
3
-3/2
-1/2
-3/2
1




0
-1
1/2
1
3
-3/2
-1/2
-3/2
1


=


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)


=


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


=


1
0
0
0
1
0
0
0
1


=I

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


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