#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 12

Hence proved  $2A^{-1}=9 I-A$

Hint:

Here, we use basic concept of determinant and inverse of matrix

$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$

Given:

$A=\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]$

Solution:

Here let’s find   $|A|, A d j(A) \& A^{-1}$

\begin{aligned} &|A|=\left|\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right|=14-12=2 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 7 & 3 \\ 4 & 2 \end{array}\right] \\ &A^{-1}=\frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right] \end{aligned}

To show  $2A^{-1}=9 I-A$

$2 A^{-1}=2 \times \frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$                          (1)

$9 I-A=\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]$                 (2)

From equation (1) and (2)

Hence,  $2A^{-1}=9 I-A$