#### Need solution for RD Sharma Maths Class 12 Chapter Algebra of Matrices Excercise 4.4 Question 1 Subquestion (iv).

Answer:$\left ( AB \right )^{T}=B^{T}A^{T}$

Given:$A=\begin{bmatrix} 2 &-3 \\ -7 & 5 \end{bmatrix},B=\begin{bmatrix} 1 &0 \\ 2 &-4 \end{bmatrix}$

Hint: The$A^{T}$ of matrix $A$ can be obtained by reflecting the elements along it’s main diagonal.

Solution:

$B^{T}=\begin{bmatrix} 1 &2 \\ 0& -4 \end{bmatrix}, A^{T}=\begin{bmatrix} 2 & -7\\ -3& 5 \end{bmatrix}$

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

$\left ( \begin{bmatrix} 2 &-3 \\ -7& 5 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 2& -4 \end{bmatrix}\right )=\begin{bmatrix} 1 &2 \\ 0& -4 \end{bmatrix}\begin{bmatrix} 2 & -7\\ -3& 5 \end{bmatrix}$

$\left ( \begin{bmatrix} 2-6 &0+12\\ -7+10& 0-20 \end{bmatrix} \right )^{T}=\begin{bmatrix} 2-6 &-7+10\\ 0+12& 0-20 \end{bmatrix}$

$\left ( \begin{bmatrix} -4 &12\\ -3& -20 \end{bmatrix} \right )^{T}=\begin{bmatrix} -4 &3\\ 12& -20 \end{bmatrix}$

$\begin{bmatrix} -4 &3\\ 12& -20 \end{bmatrix} =\begin{bmatrix} -4 &3\\ 12& -20 \end{bmatrix}$

∴LHS=RHS