Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]$

Solution:

We have $A=\left(\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right)$
To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:
$(A \mid I) \rightarrow\left(I \mid A^{-1}\right)$
Let’s proceed
$\left(\begin{array}{ccc|ccc}2 & -3 & 3 & 0 & 0 \\ 2 & 2 & 3 & 0 & 1 & 0 \\ 3 & -2 & 2 & 0 & 0 & 1\end{array}\right) \stackrel{R_{2} \rightarrow R_{2}-R_{1}}{\longrightarrow}\left(\begin{array}{ccc|ccc}2 & -3 & 3 & 1 & 0 & 0 \\ 0 & 5 & 0 & -1 & 1 & 0 \\ 3 & -2 & 2 & 0 & 0 & 1\end{array}\right)$
$\stackrel{R_{3} \rightarrow R_{3}-R_{1}}{\longrightarrow}\left(\begin{array}{ccc|ccc}2 & -3 & 3 & 1 & 0 & 0 \\ 0 & 5 & 0 & -1 & 1 & 0 \\ 1 & 1 & -1 & -1 & 0 & 1\end{array}\right)$
$\stackrel{R_{1} \rightarrow R_{1}-R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & -4 & 4 & 2 & 0 & -1 \\ 0 & 5 & 0 & -1 & 1 & 0 \\ 1 & 1 & -1 & -1 & 0 & 1\end{array}\right)$
$\stackrel{R_{3} \rightarrow R_{3}-R_{1}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & -4 & 4 & 2 & 0 & -1 \\ 0 & 5 & 0 & -1 & 1 & 0 \\ 0 & 5 & -5 & -3 & 0 & 2\end{array}\right)$
$\stackrel{R_{3} \rightarrow R_{3}-R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & -4 & 4 & 2 & 0 & -1 \\ 0 & 5 & 0 & -1 & 1 & 0 \\ 0 & 0 & -5 & -2 & -1 & 2\end{array}\right)$
$\begin{array}{l} \stackrel{R_{1} \rightarrow R_{1}+R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & 1 & 4 & 1 & 1 & -1 \\ 0 & 5 & 0 & -1 & 1 & 0 \\ 0 & 0 & -5 & -2 & -1 & 2 \end{array}\right) \\ \stackrel{R_{2} \rightarrow \frac{1}{5} R_{2} ; R_{3} \rightarrow-\frac{1}{5} R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & 1 & 4 & 1 & 1 & -1 \\ 0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{array}\right) \\ \stackrel{R_{1} \rightarrow R_{1}-R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & 0 & 4 & \frac{6}{5} & \frac{4}{5} & -1 \\ 0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{array}\right) \\ \stackrel{R_{1} \rightarrow R_{1}-4 R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{-2}{5} & 0 & \frac{3}{5} \\ 0 & 1 & 0 & -\frac{1}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1 & \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{array}\right) \end{array}$
Therefore we got the inverse of the matrix as
$A^{-1}=\left(\begin{array}{ccc} -\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5} \end{array}\right)$