#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 36 Maths Textbook Solution.

$(A B)^{-1}=\left[\begin{array}{ccc} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{array}\right]$

Hint:

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

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

Given:

$A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right], B=\left[\begin{array}{ccc} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right]$

Find  $\left ( AB \right )^{-1}$

Solution:

$A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$

So, we know that

$\left ( AB \right )^{-1}= B^{-1}A^{-1}$

So let’s find $B^{-1}$

\begin{aligned} &B^{-1}=\frac{1}{|B|} \times \operatorname{Adj}(B) \\ &|B|=1(3-0)-2(1-0)-2(2-0) \\ &|B|=3+2-4=1 \end{aligned}

Now cofactor of B

\begin{aligned} &C_{11}=3, C_{21}=2, C_{31}=6 \\ &C_{12}=1, C_{22}=1, C_{32}=2 \\ &C_{13}=2, C_{23}=2, C_{33}=5 \\ &\operatorname{Adj}(B)=\left[\begin{array}{lll} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \end{aligned}

Now,

$B^{-1}=\left[\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]$

$(A B)^{-1}=B^{-1} A^{-1}$

$=\left[\begin{array}{ccc} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]=\left[\begin{array}{ccc} 9-30+30 & -3+12-12 & 3-10+12 \\ 3-15+10 & -1+6-4 & 1-5+4 \\ 6-30+25 & -2+12-10 & 2-10+10 \end{array}\right]$

$(A B)^{-1}=\left[\begin{array}{ccc} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{array}\right]$