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If $A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} -1 & 2 &1 \\ 1 & 2 & 3 \end{bmatrix}$, then verify that $(ii) ( A - B )' = A' - B'$

Toolbox:
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where $1 \leq i \leq m$ and $1 \leq j \leq n$.
• 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.
(ii) We need to prove $(A-B)'=A'-B'$
$A'=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \rightarrow$ $A=\begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}$
$B=\begin{bmatrix} -1 & 2 &1 \\ 1 & 2 & 3 \end{bmatrix}$
$A'-B=\begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}+(-1)\begin{bmatrix} -1 & 2 &1 \\ 1 & 2 & 3 \end{bmatrix}$
$A'-B=\begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}+\begin{bmatrix} 1 &- 2 &-1 \\ -1 & -2 & -3 \end{bmatrix}$
$\;\;\;=\begin{bmatrix}3+1 &-1-2 & 0-1\\4-1 & 2-2 & 1-3\end{bmatrix} = \begin{bmatrix}4 &-3 & -1\\3 & 0 & -2\end{bmatrix}$
$\rightarrow LHS: (A-B)'=\begin{bmatrix}4 &3\\-3 & 0\\-1 & -2\end{bmatrix}$
RHS: $A'-B'=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}+(-1)\begin{bmatrix} -1 & 1 \\ 2 & 2\\1 & 3 \end{bmatrix}$
$A'-B'=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} 1 & -1 \\ -2 & -2\\-1 & -3 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 3+1 & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$
Hence LHS=RHS.