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$A=\begin{bmatrix}0 & -1 & 2\\4 & 3 & -4\end{bmatrix}\;and\;B=\begin{bmatrix}4 & 0\\1 & 3\\2 & 6\end{bmatrix},then\;verify \;that:(ii)\quad(AB)'=B'A'$

Note: This is part 2 of a 3 part question, split as 3 separate questions here.

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.
Step1:
(ii)$(AB)'=B'A'$
$A=\begin{bmatrix}0 & -1 & 2\\4 & 3 & -4\end{bmatrix}$
$B=\begin{bmatrix}4 & 0\\1 & 3\\2 & 6\end{bmatrix}$
LHS:-
$AB=\begin{bmatrix}0 & -1 & 2\\4 & 3 & -4\end{bmatrix}\begin{bmatrix}4 & 0\\1 & 3\\2 & 6\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0 & -1 & 2\\4 & 3 & -4\end{bmatrix}\begin{bmatrix}4 & 0\\1 & 3\\2 & 6\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0(4)+(-1)(1)+2(2) & 0(0)+(-1)(3)+2(6) \\4 (4)+3(1)+(-4)(2)& 4(0)+3(3)+(-4)(6) \end{bmatrix}$
$\Rightarrow \begin{bmatrix}0-1+4& 0-3+12 \\16+3-8& 0+9-24 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}3& 9\\11& -15 \end{bmatrix}$
$\Rightarrow AB= \begin{bmatrix}3& 9\\11& -15 \end{bmatrix}$
$(AB)'= \begin{bmatrix}3&11\\9& -15 \end{bmatrix}$
Step2:
RHS:-
$B'A'$
Given $B=\begin{bmatrix}4 & 0\\1 & 3\\2 & 6\end{bmatrix}$
$B'=\begin{bmatrix}4 & 1& 2\\0 & 3 & 6\end{bmatrix}$
$A=\begin{bmatrix}0 & -1& 2\\4 & 3 & -4\end{bmatrix}$
$A'=\begin{bmatrix}0 & 4\\ -1 & 3\\2 & -4\end{bmatrix}$
$B'A'=\begin{bmatrix}4 & 1 & 2\\0& 3 & 6\end{bmatrix}\begin{bmatrix}0 & 4\\-1 & 3\\2 & -4\end{bmatrix}$
$\Rightarrow \begin{bmatrix}4(0)+(1)(-1)+2(2) & 4(4)+(1)(3)+2(-4) \\0(0)+3(-1)+(6)(2)& 0(4)+3(3)+(6) (-4) \end{bmatrix}$
$\Rightarrow \begin{bmatrix}0-1+4& 16+3-8 \\0-3+12& 0+9-24 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}3 & 11\\9 & -15\end{bmatrix}$
$\Rightarrow (AB)'=B'A'$
answered Mar 23, 2013