Math 115 (Notes)

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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 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 115 (Fall 2011) web pages Created: 02 Sep 2011 Last modified: Sep 3, 2011 9:36:40 AM Comments to: dpvc@union.edu