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

Now A = IA

$\begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$

$\text{Applying R}_1\Rightarrow\frac{1}{3}\text{R}_1$

$\begin{bmatrix} 1 & 0 & \frac{-1}{3} \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$

$\text{Applying R}_2\rightarrow\text{R}_2-2\text{R}_1$

$\begin{bmatrix} 1 & 0 & \frac{-1}{3} \\ 0 & 3 & \frac{2}{3} \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$

$\text{Applying R}_3\rightarrow\text{R}_3-4\text{R}_2$

$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & \frac{1}{9} \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ \frac{8}{9} & \frac{-4}{3} & 1 \end{bmatrix}\text{A}$

$\text{Applying R}_3\rightarrow9\text{R}_3$

$\Rightarrow\begin{bmatrix} 1 & 0 & \frac{-1}{3} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 & 0 \\ \frac{-2}{9} & \frac{1}{3} & 0 \\ 8 & -12 & 9 \end{bmatrix}\text{A}$

$\text{Applying R}_1\rightarrow\text{R}_1+\frac{1}{3}\text{R}_3,\text{R}_2\rightarrow\text{R}_2-\frac{2}{9}\text{R}_3$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}\text{A}$

or I = BA

Hence, B is the inverse of A.