College Algebra, Chapter 7, 7.3, Section 7.3, Problem 22

Determine the inverse of the matrix $\left[ \begin{array}{ccc}
3 & -2 & 0 \\
5 & 1 & 1 \\
2 & -2 & 0
\end{array} \right]$ if it exists.

First, let's add the identity matrix to the right of our matrix

$\left[ \begin{array}{ccc|ccc}
3 & -2 & 0 & 1 & 0 & 0 \\
5 & 1 & 1 & 0 & 1 & 0 \\
2 & -2 & 0 & 0 & 0 & 1
\end{array} \right]$

By using Gauss-Jordan Elimination

$\displaystyle \frac{1}{3} R_1 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{1}{3} & 0 & 0 \\
5 & 1 & 1 & 0 & 1 & 0 \\
2 & -2 & 0 & 0 & 0 & 1
\end{array} \right]$

$\displaystyle R_2 - 5R_1 \to R_2 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{1}{3} & 0 & 0 \\
0 & \displaystyle \frac{13}{3} & 1 & \displaystyle \frac{-5}{3} & 1 & 0 \\
2 & -2 & 0 & 0 & 0 & 1
\end{array} \right]$

$\displaystyle R_3 - 2 R_1 \to R_3 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{1}{3} & 0 & 0 \\
0 & \displaystyle \frac{13}{3} & 1 & \displaystyle \frac{-5}{3} & 1 & 0 \\
0 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{-2}{3} & 0 & 1
\end{array} \right]$


$\displaystyle \frac{3}{13} R_2 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{1}{3} & 0 & 0 \\
0 & 1 & \displaystyle \frac{3}{13} & \displaystyle \frac{-5}{13} & \displaystyle \frac{3}{13} & 0 \\
0 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{-2}{3} & 0 & 1
\end{array} \right]$


$\displaystyle R_3 + \frac{2}{3} R_2 \to R_3 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{1}{3} & 0 & 0 \\
0 & 1 & \displaystyle \frac{3}{13} & \displaystyle \frac{-5}{13} & \displaystyle \frac{3}{13} & 0 \\
0 & 0 & \displaystyle \frac{2}{13} & \displaystyle \frac{-12}{13} & \displaystyle \frac{2}{13} & 1
\end{array} \right]$

$\displaystyle \frac{13}{2} R_3 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{1}{3} & 0 & 0 \\
0 & 1 & \displaystyle \frac{3}{13} & \displaystyle \frac{-5}{13} & \displaystyle \frac{3}{13} & 0 \\
0 & 0 & 1 & -6 & 1 & \displaystyle \frac{13}{2}
\end{array} \right]$

$\displaystyle R_2 - \frac{3}{13} R_3 \to R_2 $

$\left[ \begin{array}{ccc|ccc}
1 & \displaystyle \frac{-2}{3} & 0 & \displaystyle \frac{1}{3} & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & \displaystyle \frac{-3}{2} \\
0 & 0 & 1 & -6 & 1 & \displaystyle \frac{13}{2}
\end{array} \right]$

$\displaystyle R_1 + \frac{2}{3} R_2 \to R_1 $

$\left[ \begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & 0 & -1 \\
0 & 1 & 0 & 1 & 0 & \displaystyle \frac{-3}{2} \\
0 & 0 & 1 & -6 & 1 & \displaystyle \frac{13}{2}
\end{array} \right]$


The inverse matrix can now be found in the right half of our reduced row-echelon matrix. So the inverse matrix is

$\left[ \begin{array}{ccc}
1 & 0 & -1 \\
1 & 0 & \displaystyle \frac{-3}{2} \\
-6 & 1 & \displaystyle \frac{13}{2}
\end{array} \right]$

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