#### Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 32

$A^{-1}=\left[\begin{array}{cc} -16 & 3 \\ 24 & -5 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{rr} -1 & 1 \\ -2 & 1 \end{array}\right], C=\left[\begin{array}{cc} 2 & -1 \\ 0 & 4 \end{array}\right]$

Solution:

Then the given equation becomes as,

\begin{aligned} &A X B=C \\ &X=A^{-1} C B^{-1} \\ &|A|=15-14=1 \\ &|B|=-1+2=1 \end{aligned}

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

\begin{aligned} B^{-1}=\frac{1}{|B|} \times \operatorname{Adj}(B) &=1\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right] \end{aligned}

\begin{aligned} &X=A^{-1} C B^{-1} \\ &=1\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 0 & 4 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} 10+0 & -5-8 \\ -14+0 & 7+12 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right] \end{aligned}

\begin{aligned} &=\left[\begin{array}{cc} 10-26 & -10+13 \\ -14+38 & 14-19 \end{array}\right] \\ &X=\left[\begin{array}{cc} -16 & 3 \\ 24 & -5 \end{array}\right] \end{aligned}