#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 14

$A^{-1}=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{lc} a & b \\ c & \frac{1+b c}{a} \end{array}\right]$

Solution:

\begin{aligned} &|A|=\frac{a+a b c}{a}-b c \\ &=\frac{a+a b c-a b c}{a}=\frac{a}{a}=1 \end{aligned}

So, hence $A^{-1}$  exist

Cofactor of A

\begin{aligned} &C_{11}=\frac{1+b c}{a}, C_{12}=-c \\\\ &C_{21}=-b, C_{22}=a \\\\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} \frac{1+b c}{a} & -c \\ -b & a \end{array}\right]^{T}=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\\\ &=\frac{1}{1}\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \\\\ &A^{-1}=\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right] \end{aligned}

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

LHS

$a A^{-1}=a\left[\begin{array}{cc} \frac{1+b c}{a} & -b \\ -c & a \end{array}\right]=\left[\begin{array}{cc} 1+b c & -a b \\ -a c & a^{2} \end{array}\right]$

RHS

\begin{aligned} &\left(a^{2}+b c+1\right) I-a A \\\\ &=\left[\begin{array}{cc} a^{2}+b c+1 & 0 \\ 0 & a^{2}+b c+1 \end{array}\right]-\left[\begin{array}{cc} a^{2} & a b \\ a c & 1+b c \end{array}\right]\\ \\ &=\left[\begin{array}{cc} 1+b c & -a b \\ -a c & a^{2} \end{array}\right] \end{aligned}

LHS = RHS