#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 17 Maths Textbook Solution.

$\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]$

Hint:

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

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

Given:

\begin{aligned} &A=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right] \\\\ &A^{2}-4 A+I=0 \\\\ &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \& O=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}

Solution:

\begin{aligned} &A^{2}=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 4+3 & 6+6 \\ 2+2 & 3+4 \end{array}\right] \\\\ &A^{2}=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right] \end{aligned}

\begin{aligned} &4 A=4 \times\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 8 & 12 \\ 4 & 8 \end{array}\right] \\ \\&I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}

\begin{aligned} &A^{2}-4 A+I=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]-\left[\begin{array}{cc} 8 & 12 \\ 4 & 8 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \\&=\left[\begin{array}{cc} 7-8+1 & 12-12+0 \\ 4-4+0 & 7-8+1 \end{array}\right]=\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}

Hence

\begin{aligned} &A^{2}-4 A+I=0 \\ &A \times A-4 A=-I \end{aligned}

Let’s multiply $A^{-1}$  both side

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