#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 20

$X = 9 \: and\: y =14$  ,   $A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 5 & -3 \\ -2 & 4 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]$

Solution:

\begin{aligned} &A^{2}=\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{cc} 16+6 & 12+15 \\ 8+10 & 6+25 \end{array}\right]=\left[\begin{array}{cc} 22 & 27 \\ 18 & 31 \end{array}\right] \\\\ &A^{2}-x A+y I=\left[\begin{array}{ll} 22 & 37 \\ 18 & 31 \end{array}\right]-x\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 0 & y \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}

\begin{aligned} &22-4 x+y=0 \\ &4 x-y=22 \\ &x=9 \end{aligned}

So,

\begin{aligned} &4 x-y=22 \\ &4 \times 9-y=22 \\ &36-22=y \\ &y=14 \end{aligned}

So  $A^{2}-9 A+14 I=0$

Multiply by $A^{-1}$  both sides,

\begin{aligned} &A \times A \times A^{-1}-9 A \times A^{-1}+14 I \times A^{-1}=0 \\\\ &A-9 I+14 A^{-1}=0 \\\\ &A^{-1}=\frac{1}{14}[9 I-A] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{14}\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right] \\ \\&A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 5 & -3 \\ -2 & 4 \end{array}\right] \end{aligned}