# Find the inverse of each of the matrices, if it exists.

Question:

Find the inverse of each of the matrices, if it exists.

$\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$

Solution:

Let $A=\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$

We know that A = IA

$\therefore\left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$

$\Rightarrow\left[\begin{array}{rrr}2 & -3 & 3 \\ 0 & 5 & 0 \\ 3 & -2 & 2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A \quad\left(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\right)$

$\Rightarrow\left[\begin{array}{ccc}2 & -3 & 3 \\ 0 & 1 & 0 \\ 3 & -2 & 2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1\end{array}\right] A \quad\left(\mathrm{R}_{2} \rightarrow \frac{1}{5}\right)$

$\Rightarrow\left[\begin{array}{ccc}-1 & -1 & 1 \\ 0 & 1 & 0 \\ 3 & -2 & 2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & -1 \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ 0 & 0 & 1\end{array}\right] A \quad\left(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\right)$

$\Rightarrow\left[\begin{array}{ccc}-1 & 0 & 1 \\ 0 & 1 & 0 \\ 3 & 0 & 2\end{array}\right]=\left[\begin{array}{ccc}\frac{4}{5} & \frac{1}{5} & -1 \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ -\frac{2}{5} & \frac{2}{5} & 1\end{array}\right] A \quad\left(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}\right.$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}+2 \mathrm{R}_{2}\right)$

$\Rightarrow\left[\begin{array}{lll}-1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 5\end{array}\right]=\left[\begin{array}{lll}\frac{4}{5} & \frac{1}{5} & -1 \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ 2 & 1 & -2\end{array}\right] A \quad\left(R_{3} \rightarrow R_{3}+3 R_{1}\right)$

$\Rightarrow\left[\begin{array}{ccc}-1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}\frac{4}{5} & \frac{1}{5} & -1 \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right] A \quad\left(\mathrm{R}_{3} \rightarrow \frac{1}{5} \mathrm{R}_{3}\right)$

$\Rightarrow\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}\frac{2}{5} & 0 & -\frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right] A \quad\left(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\right)$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right] A \quad\left(\mathrm{R}_{1} \rightarrow(-1) \mathrm{R}_{1}\right)$

$\therefore A^{-1}=\left[\begin{array}{ccc}-\frac{2}{5} & 0 & \frac{3}{5} \\ -\frac{1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]$