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

Question:

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

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

Solution:

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

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

We know that A = IA

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

Applying $R_{1} \rightarrow \frac{1}{2} R_{1}$, we have:

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

Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-5 \mathrm{R}_{1}$, we have:

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

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

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

Applying $\mathrm{R}_{3} \rightarrow 2 \mathrm{R}_{3}$, we have:

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

Applying $R_{1} \rightarrow R_{1}+\frac{1}{2} R_{3}$, and $R_{2} \rightarrow R_{2}-\frac{5}{2} R_{3}$, we have:

$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lrl}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] A$

$\therefore A^{-1}=\left[\begin{array}{lrl}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$