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#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 7 Subquestion (iii)

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

Hint:

Here, we use basic concept of inverse.

Given:

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

Solution:

We know that

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

Let’s find  $Adj\left ( A \right )$

\begin{aligned} &C_{11}=\frac{1+b c}{a} \\ &C_{12}=-c \\ &C_{21}=-b \\ &C_{22}=a \\ &C_{i j}=\left[\begin{array}{cc} \frac{1+b c}{a} & -c \\ -b & a \end{array}\right] \end{aligned}

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

So, let’s put value in formula

\begin{aligned} A^{-1} &=\frac{1}{1} \times\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}