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# Let $A=\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix},B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix},C=\begin{bmatrix}2 & 0\\1 & 2\end{bmatrix}\;$and$\;a=4,b=-2$.Show that$\;(g)\;(AB)^T=B^TA^T.$

Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Step1:
(g)Given:
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\;B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
LHS
$AB=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1(4)+2(1) &1(0)+2(5)\\-1(4)+3(1) & -1(0)+3(5)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}4+2 &0+10\\-4+3 & 0+15\end{bmatrix}$
$\;\;\;=\begin{bmatrix}6 &10\\-1 & 15\end{bmatrix}$
$(AB)^T=\begin{bmatrix} 6 & -1\\10 & 15\end{bmatrix}$
Step2:
RHS:-
$B^TA^T$
$B=\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}$
$B^T=\begin{bmatrix}4 & 1\\0 & 5\end{bmatrix}$
$A=\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$A^T=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$
$B^TA^T=\begin{bmatrix}4 & 1\\0 & 5\end{bmatrix}\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$
$\qquad\quad=\begin{bmatrix}4(1)+1(2) & 4(-1)+1(3)\\0(1)+5(2) & 0(-1)+5(3)\end{bmatrix}$
$\qquad\quad=\begin{bmatrix}4+2 & -4+3\\0+10& 0+15\end{bmatrix}$
$\qquad\quad=\begin{bmatrix}6 & -1\\10& 15\end{bmatrix}$
$\Rightarrow LHS=RHS.$
$\Rightarrow (AB)^T=B^TA^T$