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# Given $A=\begin{bmatrix}2 & 4 & 0\\3 & 9 & 6\end{bmatrix}\;and\;B=\begin{bmatrix}1& 4\\2 & 8\\1 & 3\end{bmatrix}.\;Is \;(AB)'=B'A'?$

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:
By the property of transpose of a matrix, we have $(AB)'=B'A'$
Given:
$A=\begin{bmatrix}2 & 4 & 0\\3 & 9 & 6\end{bmatrix}_{2\times 3}\;\;B=\begin{bmatrix}1& 4\\2 & 8\\1 & 3\end{bmatrix}_{3\times 2}$
$AB=\begin{bmatrix}2 & 4 & 0\\3 & 9 & 6\end{bmatrix}\begin{bmatrix}1& 4\\2 & 8\\1 & 3\end{bmatrix}$
$AB=\begin{bmatrix}2 (1)+4(2)+0(1)& 2(4)+4(8)+0(3)\\3(1)+9(2)+6(1) & 3(4)+9(8)+6(3)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2 +8+0& 8+32+0\\3+18+6 & 12+72+18\end{bmatrix}$
$\;\;\;=\begin{bmatrix}10& 40\\27 & 102\end{bmatrix}$
$(AB)'=\begin{bmatrix}10& 27\\40 & 102\end{bmatrix}$ [Transpose of a matrix can be obtained by changing the rows and a column.]
Step2:
Let $B'A'$
Given:B=$\begin{bmatrix}1 & 4\\2 & 8\\1 & 3\end{bmatrix}$
$B'=\begin{bmatrix}1 & 2 &1\\4 & 8 & 3\end{bmatrix}$
$A=\begin{bmatrix}2 & 4 & 0\\3 & 9 & 6\end{bmatrix}$
$A'=\begin{bmatrix}2 & 3 \\4 & 9 \\0 & 6\end{bmatrix}$
$B'A'=\begin{bmatrix}1 & 2 & 1\\4 & 8 & 3\end{bmatrix}\begin{bmatrix}2& 3\\4 & 9\\0 & 6\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1 (2)+2(4)+1(0)& 1(3)+2(9)+1(6)\\4(2)+8(4)+3(0) & 4(3)+8(9)+3(6)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2 +8+0& 3+18+6\\ 8+32+0& 12+72+18\end{bmatrix}$
$\;\;\;=\begin{bmatrix}10& 27\\40 & 102\end{bmatrix}$
$\Rightarrow(AB)'=B'A'$