Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

We know A = IA

$\Rightarrow\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\text{A}$

$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$

$\Big[\text{Applying R}_1\rightarrow\frac{1}{2}\text{R}_1\Big]$

$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$

$\big[\text{Applying R}_2\rightarrow\text{R}_2-5\text{R}_1\big]$

$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2}\\ 0 & 0 &\frac{1}{2} \end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1\end{bmatrix}\text{A}$

$\big[\text{Applying R}_3\rightarrow\text{R}_3-\text{R}_2\big]$

$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2}\\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix}\frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 5 & -2 & 2\end{bmatrix}\text{A}$

$\big[\text{Applying R}_3\rightarrow\ 2\text{R}_3\big]$

$\Rightarrow\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\text{A}$

$\Big[\text{Applying R}_1\rightarrow\text{R}_1+\frac{1}{2}\text{ R}_3\text{ and R}_2\rightarrow\text{R}_2-\frac{5}{2}\text{R}_3\Big]$

$\Rightarrow\text{A}^{-1}=\begin{bmatrix} 3 & -1 & 1\\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$