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

Now, A = IA

$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{A}$

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

$\begin{bmatrix} 1 & \frac{10}{7} \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}\text{A}$

Applying R₂ → R₂ - 2R₁

$\begin{bmatrix} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ \frac{-2}{3} & 1 \end{bmatrix}\text{A}$

Applying R₂ → 3R₂

$\begin{bmatrix}1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ -2 & 3 \end{bmatrix}\text{A}$

$\text{Applying R}_1\rightarrow\text{R}_1-\frac{10}{3}\text{ R}_3$

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\text{A}$

I = BA

Hence, B is the inverse of A.