#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 13

Hence proved  $A-3 I=2\left(I+3 A^{-1}\right)$

Hint:

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

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

Given:

$A=\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]$

Solution:

Let’s find  $|A|, A d j(A) \& A^{-1}$

\begin{aligned} &|A|=4-10=-6 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -5 \\ -2 & 4 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{-6}\left[\begin{array}{cc} 1 & -5 \\ -2 & 4 \end{array}\right] \end{aligned}

To show  $A-3 I=2\left(I+3 A^{-1}\right)$

LHS

$A-3 I=\left[\begin{array}{ll} 4 & 5 \\ 2 & 1 \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 5 \\ 2 & -2 \end{array}\right]$                        (1)

RHS

\begin{aligned} &2\left(I+3 A^{-1}\right)=2 I+6 A^{-1}=2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\frac{6}{6}\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right] \\ &=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]+\left[\begin{array}{cc} -1 & 5 \\ 2 & -4 \end{array}\right] \end{aligned}

$= \left[\begin{array}{cc} 1 & 5 \\ 2 & -2 \end{array}\right]$                               (2)

Here from equation (1) and (2)

$A-3 I=2\left(I+3 A^{-1}\right)$