Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 10 Suquestion (ii) 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:

$A=\left[\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]$

Solution:

Let’s find $\left | A \right |Adj\left ( A \right ) \& \: A^{-1}$

$|A|=\left|\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right|=6-5=1$

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

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

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

$|B|=\left|\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right|=16-15=1$

$\operatorname{Adj}(B)=\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right]$

$B^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)=\frac{1}{1}\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right]$

Then find AB

$A \times B=\left[\begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right]\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 11 & 14 \\ 29 & 37 \end{array}\right]$

Let’s find   $\left | AB \right |,Adj\left ( A \right ) \& \:\left ( AB \right ) ^{-1}$

$|A B|=407-406=1$

$\operatorname{Adj}(A B)=\left[\begin{array}{cc} 37 & -14 \\ -29 & 11 \end{array}\right]$

$(A B)^{-1}=\frac{1}{|A B|} \times \operatorname{Adj}(A)=\left[\begin{array}{cc} 37 & -14 \\ -29 & 11 \end{array}\right]$                            (1)

Now

\begin{aligned} &B^{-1} A^{-1}=\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -1 \\ -5 & 2 \end{array}\right] \\ & \end{aligned}

$B^{-1} A^{-1}=\left[\begin{array}{cc} 37 & -14 \\ -29 & 11 \end{array}\right]$                         (2)

Hence proved from equation (1) and (2)

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