#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 11

$\left[\begin{array}{cc} -47 & \frac{39}{2} \\ 41 & -17 \end{array}\right]$

Hint:

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

$A^{-1}=\frac{1}{|A|} \times A d j(A)$

Given:

$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right]$

Solution:

$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right]$

Let’s find $\left | A \right |$

\begin{aligned} &|A|=\left|\begin{array}{cc} 3 & 2 \\ 7 & 5 \end{array}\right|=15-14=1 \\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]=\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \end{aligned}

\begin{aligned} &B=\left[\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right] \\ &|B|=54-56=-2 \\ &\operatorname{Adj}(B)=\left[\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right] \\ &B^{-1}=\frac{1}{|B|} \times \operatorname{Adj}(A)=\frac{1}{-2}\left[\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right] \end{aligned}

Now, we know that

\begin{aligned} &(A B)^{-1}=B^{-1} A^{-1} \\ &=\frac{1}{-2}\left[\begin{array}{cc} 9 & -7 \\ -8 & 6 \end{array}\right]\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right] \\ &=\frac{1}{-2}\left[\begin{array}{cc} 45+49 & -18-21 \\ -40-42 & 16+18 \end{array}\right] \end{aligned}

\begin{aligned} &=-\frac{1}{2}\left[\begin{array}{cc} 94 & -39 \\ -82 & 34 \end{array}\right] \\ &(A B)^{-1}=\left[\begin{array}{cc} -47 & \frac{39}{2} \\ 41 & -17 \end{array}\right] \end{aligned}