#### Need solution for RD Sharma Maths Class 12 Chapter Algebra of Matrices Excercise 4.4 Question 6 Subquestion (ii).

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

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

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

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

Solution:

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

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

$\left ( \begin{bmatrix} 1 &3 \\ 2& 4 \end{bmatrix}\begin{bmatrix} 1 &4 \\ 2 & 5 \end{bmatrix} \right )^{T}=\begin{bmatrix} 1 &2 \\ 4& 5 \end{bmatrix}\begin{bmatrix} 1 &2 \\ 3& 4 \end{bmatrix}$

$\begin{bmatrix} 1+6 &4+15 \\ 2+8 & 8+20 \end{bmatrix}^{T}=\begin{bmatrix} 1+6 & 2+8\\ 4+15 &8+20 \end{bmatrix}$

$\begin{bmatrix} 7 & 19\\ 10 & 28 \end{bmatrix}^{T}=\begin{bmatrix} 7 &10 \\ 19& 28 \end{bmatrix}$

$\begin{bmatrix} 7 & 19\\ 10 & 28 \end{bmatrix}=\begin{bmatrix} 7 &19 \\ 19& 28 \end{bmatrix}$

∴LHS=RHS

Hence, $\left (AB \right )^{T}=B^{T}A^{T}$ is proved.