## Using the Augmented Matrix to Find the Inverse

Suppose we want to find the inverse of a matrix. There are several methods to do this. One method is using 'the augmented matrix' and row reducing. Example: Let
$A = \left( \begin{array}{ccc} 2 & 1 & 0 \\1 & 3 & 4 \\ 1 & 1 & 2 \end{array} \right)$

The augmented matrix is obtained by adding extra columns to the right. The columns together make an identity matrix.. We get
$\left( \begin{array}{ccc} 2 & 1 & 0 \\1 & 3 & 4 \\ 1 & 1 & 2 \end{array} \right| \left| \begin{array}{ccc} 1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$

We want to times or divide rows by numbers, and add or subtract multiples of rows from each other so the left hand side of the vertical line becomes the identity. The right hand side then becomes the inverse. We always start to make the leading 1, in the upper left corner a 1, and add or subtract multiples of the first row so other entries in the first column become zero.
Divide row 1 by 2
$\left( \begin{array}{ccc} 1 & 0.5 & 0 \\ 1 & 3 & 4 \\ 1 & 1 & 2 \end{array} \right| \left| \begin{array}{ccc} 0.5 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$

Subtract row 1 from first and second rows.
$\left( \begin{array}{ccc} 1 & 0.5 & 0 \\ 0 & 2.5 & 4 \\ 0 & 0.5 & 2 \end{array} \right| \left| \begin{array}{ccc} 0.5 & 0 & 0 \\ -0.5 & 1 & 0 \\ -0.5 & 0 & 1 \end{array} \right)$

Divide the second row by 2.5
$\left( \begin{array}{ccc} 1 & 0.5 & 0 \\ 0 & 1 & 1.6 \\ 0 & 0.5 & 2 \end{array} \right| \left| \begin{array}{ccc} 0.5 & 0 & 0 \\ -0.2 & 0.4 & 0 \\ -0.5 & 0 & 1 \end{array} \right)$

Subtract half the second row from both the first and third rows.
$\left( \begin{array}{ccc} 1 & 0 & -0.8 \\ 0 & 1 & 1.6 \\ 0 & 0 & 1.2 \end{array} \right| \left| \begin{array}{ccc} 0.6 & -0.2 & 0 \\ -0.2 & 0.4 & 0 \\ -0.4 & -0.2 & 1 \end{array} \right)$

Divide the third row by 1.2
$\left( \begin{array}{ccc} 1 & 0 & -0.8 \\ 0 & 1 & 1.6 \\ 0 & 0 & 1 \end{array} \right| \left| \begin{array}{ccc} 0.6 & -0.2 & 0 \\ -0.2 & 0.4 & 0 \\ -1/3 & -1/6 & 5/6 \end{array} \right)$

Add 0.8 times the third row to the first row and subtract 1.6 times the third row from the second row.
$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right| \left| \begin{array}{ccc} 1/3 & -1/3 & 2/3 \\ 1/3 & 2/3 & -4/3 \\ -1/3 & -1/6 & 5/6 \end{array} \right)$

Then
$A^{-1} = \left( \begin{array}{ccc} 1/3 & -1/3 & 2/3 \\ 1/3 & 2/3 & -4/3 \\ -1/3 & -1/6 & 5/6 \end{array} \right)$