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# For the matrices $A$ and $B$, verify that $(AB)' = B'A'$ , where $$\text{ (i) } A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \text{ , } B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} \qquad$$

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Toolbox:
• 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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• 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.
Given:
$A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$
$B=\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$
LHS:
AB=$\begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$
$\;\;\;AB=\begin{bmatrix} -1\times -1 & 1\times 2 & 1\times 1\\-4\times -1 & -4\times 2 & -4\times 1\\3\times -1 & 3\times 2 & 3\times 1 \end{bmatrix}\Rightarrow \begin{bmatrix}-1 &2 & 1\\4 & -8 & -4\\-3 & 6 & 3\end{bmatrix}$
$(AB)'=\begin{bmatrix}-1 & 4 & -3\\2 & -8 & 6\\1 &-4 &3\end{bmatrix}$
RHS:
B'A'
Given $B=\begin{bmatrix}-1 & 2 & 1\end{bmatrix}$
$B'=\begin{bmatrix}-1\\2\\1\end{bmatrix}$
$A=\begin{bmatrix}1\\-4\\3\end{bmatrix}$
$A'=\begin{bmatrix}1 & -4 & 3\end{bmatrix}$
$B'A'=\begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}1 & -4 & 3\end{bmatrix}$
$\;\;\;=\begin{bmatrix} -1\times 1 & -1\times -4 & -1\times 3\\2\times 1 & 2\times -4 & 2\times 3\\1\times 1 & 1\times -4 & 1\times 3 \end{bmatrix}$
$B'A'=\begin{bmatrix}-1 & 4 & -3\\2 & -8 & 6\\1 &-4 &3\end{bmatrix}$
Hence LHS=RHS.