#### Need solution for RD Sharma Maths Class 12 Chapter Algebra of Matrices Excercise 4.4 Question 3 Subquestion (ii).

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

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

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:

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

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

$\left ( \begin{bmatrix} 1 & -1 &0 \\ 2 & 1 &1 \\ 3 &3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3\\ 2 & 1 & 3\\ 0 & 1 & 1 \end{bmatrix}\right )^{T}=\begin{bmatrix} 1 & 2 &0 \\ 2 & 1 & 1\\ 3 & 3 &1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 1\\ -1 & 1 & 2\\ 0 &3 & 1 \end{bmatrix}$

$\begin{bmatrix} 1-2+0 & 2-1+0 &3-3+0 \\ 2+2+0 & 4+1+3 &6+3+3 \\ 1+4+0 & 2+2+1 & 3+6+1 \end{bmatrix}^{T}=\begin{bmatrix} 1-2+0 & 2+2+0 &1+4+0 \\ 2-1+0 &4+1+3 &2+2+1 \\ 3-3+0 &6+3+3 & 3+6+1 \end{bmatrix}$

$\begin{bmatrix} -1 & 1 &0 \\ 4 & 8 &12\\ 5 & 5 & 10 \end{bmatrix}^{T}=\begin{bmatrix} -1 & 4 &5 \\ 1 &8 &5 \\ 0&12 & 10 \end{bmatrix}$

$\begin{bmatrix} -1 & 4 &5 \\ 1 &8 &5 \\ 0&12 & 10 \end{bmatrix}=\begin{bmatrix} -1 & 4 &5 \\ 1 &8 &5 \\ 0&12 & 10 \end{bmatrix}$

∴LHS=RHS