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

Question:

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

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

Solution:

Let $A=\left[\begin{array}{llr}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]$

We know that A = IA

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

Applying $R_{2} \rightarrow R_{2}+3 R_{1}$ and $R_{3} \rightarrow R_{3}-2 R_{1}$, we have:

$\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] A$

Applying $R_{1} \rightarrow R_{1}+3 R_{3}$ and $R_{2} \rightarrow R_{2}+8 R_{3}$, we have:

$\left[\begin{array}{lll}1 & 0 & 10 \\ 0 & 1 & 21 \\ 0 & -1 & 4\end{array}\right]=\left[\begin{array}{lll}-5 & 0 & 3 \\ -13 & 1 & 8 \\ -2 & 0 & 1\end{array}\right] A$

Applying $R_{3} \rightarrow R_{3}+R_{2}$, we have:

$\left[\begin{array}{lll}1 & 0 & 10 \\ 0 & 1 & 21 \\ 0 & 0 & 25\end{array}\right]=\left[\begin{array}{lll}-5 & 0 & 3 \\ -13 & 1 & 8 \\ -15 & 1 & 9\end{array}\right] A$

Applying $\mathrm{R}_{3} \rightarrow \frac{1}{25} \mathrm{R}_{3}$, we have:

$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-5 & 0 & 3 \\ -13 & 1 & 8 \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right] A$

Applying $R_{1} \rightarrow R_{1}-10 R_{3}$, and $R_{2} \rightarrow R_{2}-21 R_{3}$, we have:

$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right] A$

$\therefore A^{-1}=\left[\begin{array}{ccc}1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right]$