Up: Math 15 Selected Class 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.


Up: Math 15 Selected Class Notes

Comments to: dpvc@union.edu
Created: Feb 9 1998 --- Last modified: Nov 11, 1998 10:43:57 AM