Math 15H (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 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 = \Matrix{ 2& 0& -4\cr 2& 1& -2\cr 1& 1& 0\cr }

Solution:

\Matrix{ 2& 0& -4& & 1& 0& 0\cr 3& 1& -2& \|& 0& 1& 0\cr 1& 1& 0& & 0& 0& 1\cr }		form the augmented matrix [\,A\mid I\;]
\Matrix{ 1& 1& 0& & 0& 0& 1\cr 3& 1& -2& \|& 0& 1& 0\cr 2& 0& -4& & 1& 0& 0\cr }		Swap rows 1 and 3 (to get a 1 at the upper left)
\Matrix{ 1& 1& 0& & 0& 0& 1\cr 0& -2& -2& \|& 0& 1& -3\cr 0& -2& -4& & 1& 0& -2\cr }		Add -3 times row 1 to row 2 Add -2 times row 1 to row 3
\Matrix{ 1& 1& 0& & 0& \-0& 1\cr 0& 1& 1& \|& 0& -1/2& 3/2\cr 0& -2& -4& & 1& \-0& -2\cr }		Multiply row 2 by -1/2
\Matrix{ 1& 0& -1& & 0&\-1/2& -1/2\cr 0& 1& 1& \|& 0& -1/2&\-3/2\cr 0& 0& -2& & 1& -1& \-1\cr }		Add -1 times row 2 to row 1 Add 2 times row 2 to row 3
\Matrix{ 1& 0& -1& & 0&\-1/2& -1/2\cr 0& 1& 1& \|& 0& -1/2&\-3/2\cr 0& 0& 1& & -1/2&\-1/2& -1/2\cr }		Multiply row 3 by -1/2
\Matrix{ 1& 0& 0& & -1/2& \-1& -1\cr 0& 1& 0& \|&\-1/2& -1& \-2\cr 0& 0& 1& & -1/2& 1/2& -1/2\cr }		Add row 3 to row 1 Add -1 times row 3 to row 2

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

\displaystyle A^{-1} = \Matrix{ -1/2& \-1& -1\cr \-1/2& -1& \-2\cr -1/2& 1/2& -1/2\cr } = {1\over 2} \Matrix { -1&\-2& -2\cr \-1& -2&\-4\cr -1&\-1& -1\cr }

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

\begin{eqnarray} A^{-1}A &=& \displaystyle {1\over 2} \Matrix { -1&\-2& -2\cr \-1& -2&\-4\cr -1&\-1& -1\cr } \Matrix{ 2& 0& -4\cr 2& 1& -2\cr 1& 1& 0\cr }\cr &\cr &=& \displaystyle {1\over 2}\Matrix{ (-1)\.2+2\.3+(-2)\.1& (-1)\.0+2\.1+(-2)\.1& (-1)(-4)+2(-2)+(-2)\.0\cr 1\.2+(-2)\.3+4\.1& 1\.0+(-2)\.1+4\.1& 1(-4)+(-2)(-2)+4\.0\cr (-1)\.2+1\.3+(-1)\.1& (-1)\.0+1\.1+(-1)\.1& (-1)(-4)+1\.(-2)+(-1)\.0\cr }\cr &\cr &=& \displaystyle {1\over 2}\Matrix{ -2+6-2& 0+2-2& \-4-4+0\cr \-2-6+4& 0-2+4& -4+4+0\cr -2+3-1& 0+1-1& \-4-2+0\cr }\cr &\cr &=& \displaystyle {1\over 2}\Matrix{ 2& 0& 0\cr 0& 2& 0\cr 0& 0& 2\cr } = \Matrix{1& 0& 0\cr 0& 1& 0\cr 0& 0& 1\cr} = I\cr \end{eqnarray}

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

Math 15H (Fall 2004) web pages Created: 04 Sep 2004 Last modified: 13 Nov 2004 09:50:38 Comments to: dpvc@union.edu

Selected Course Notes Gauss-Jordan Elimination Notes