Find the inverse of the following matrices by using elementary row transformations: $\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

Solution:

For elementary row operation we have,
$\begin{array}{l} A=I A \\ \Rightarrow\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A} \end{array}$
Applying $r_{1} \rightarrow \frac{1}{2} r_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $r_{2} \rightarrow r_{2}-5 r_{1}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A$
Applying $r_{3} \rightarrow r_{3}-r_{2}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{1}{2} & -1 & 1 \end{array}\right] \mathrm{A}$
Applying $r_{3} \rightarrow 2 r_{3}$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 5 & -2 & 2 \end{array}\right] A$
Applying $r_{1} \rightarrow r_{1}+\frac{1}{2} r_{3}$ and $r_{2} \rightarrow r_{2}-\frac{5}{2} r_{3}$
$\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] A$
Therefore, as it is known that
$\mathrm{I}=\mathrm{A}^{-1} \mathrm{~A}$
Therefore
$\Rightarrow A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$