#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 10 Suquestion (i) Maths Textbook Solution.

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

Hint:

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

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

Given:

$A=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right], B=\left[\begin{array}{ll} 4 & 6 \\ 3 & 2 \end{array}\right]$

Solution:

Let’s find  $\left | A \right |$

$|A|=\left|\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right|=15-14=1$

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

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

$A^{-1}=\frac{1}{1}\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]$

Then let’s find  $\left | B \right |Adj\left ( B \right ) \& \: B^{-1}$

$|B|=\left|\begin{array}{cc} 4 & 6 \\ 3 & 2 \end{array}\right|=8-18=-10$

$\operatorname{Adj}(B)=\left[\begin{array}{cc} 2 & -6 \\ -3 & 4 \end{array}\right]$

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

$B^{-1}=\frac{1}{-10} \times\left[\begin{array}{cc} 2 & -6 \\ -3 & 4 \end{array}\right]$

Then find $AB$

\begin{aligned} &A \times B=\left[\begin{array}{ll} 3 & 2 \\ 7 & 5 \end{array}\right] \times\left[\begin{array}{ll} 4 & 6 \\ 3 & 2 \end{array}\right] \\ &\end{aligned}

$=\left[\begin{array}{cc} 12+6 & 18+4 \\ 28+15 & 42+10 \end{array}\right]=\left[\begin{array}{ll} 18 & 22 \\ 43 & 52 \end{array}\right]$

Then let’s find  $\left | AB \right |,Adj\left ( AB \right )$  and inverse of  $AB$

$|A B|=\left|\begin{array}{cc} 18 & 22 \\ 43 & 52 \end{array}\right|=936-946=-10$

$\operatorname{Adj}(A B)=\left[\begin{array}{cc} 52 & -22 \\ -43 & 18 \end{array}\right]$

$(A B)^{-1}=\frac{1}{-10} \times\left[\begin{array}{cc} 52 & -22 \\ -43 & 18 \end{array}\right]=\frac{1}{10}\left[\begin{array}{cc} -52 & 22 \\ 43 & -18 \end{array}\right]$                 (1)

Now $B^{-1}A^{-1}$

$=\frac{1}{10}\left[\begin{array}{cc} 2 & -6 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]$

$=\frac{1}{10}\left[\begin{array}{cc} 10+42 & -4-18 \\ -15-28 & 6+12 \end{array}\right]$

$=\frac{1}{10}\left[\begin{array}{cc} -52 & 22 \\ 43 & -18 \end{array}\right]$                           (2)

From equation (1) and (2)

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

Hence proved