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# If $A=\begin{bmatrix}1 & 2\\4 & 1\\5 & 0\end{bmatrix},B=\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix},then\;verify\;that:(ii)\quad(A-B)'=A'-B'$.

Note: This is part 2 of a 2 part question, split as 2 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.
• 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 ≤ i ≤ m and 1 ≤ j ≤ n.
Step1:
$A=\begin{bmatrix}1 & 2\\4 & 1\\5 & 6\end{bmatrix}$
$B=\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix}$
LHS:-
$(A-B)'$
$(A-B)=\begin{bmatrix}1 & 2\\4 & 1\\5 & 6\end{bmatrix}-\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}1-1 & 2-2\\4-6 & 1-4\\5-7 & 6-3\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}0 & 0\\-2 & -3\\-2 & 3\end{bmatrix}$
$(A-B)'=\begin{bmatrix}0 & -2 & -2\\0 & -3 & 3\end{bmatrix}$
Step2:
RHS:-
$A'-B'$
$A=\begin{bmatrix}1 & 2\\4 & 1\\5 & 6\end{bmatrix}$
$A'=\begin{bmatrix}1 & 4 & 5\\2 &1& 6\end{bmatrix}$
$B=\begin{bmatrix}1 & 2\\6 & 4\\7 & 3\end{bmatrix}$
$B'=\begin{bmatrix}1 & 6 & 7\\2 & 4 & 3\end{bmatrix}$
$A'-B'=\begin{bmatrix}1 & 4 & 5\\2 &1& 6\end{bmatrix}-\begin{bmatrix}1 & 6 & 7\\2 & 4 & 3\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}1 & 4 & 5\\2 &1& 6\end{bmatrix}+(-1)\begin{bmatrix}1 & 6 & 7\\2 & 4 & 3\end{bmatrix}$
$\;\;\;\;\;=\begin{bmatrix}1 & 4 & 5\\2 &1& 6\end{bmatrix}+\begin{bmatrix}-1 & -6 & -7\\-2 &- 4 & -3\end{bmatrix}$
$\;\;\;\;\;\;\;=\begin{bmatrix}1-1 & 4-6 & 5-7\\2-2 & 1-4 & 6-3\end{bmatrix}$
$\;\;\;\;\;\;\;=\begin{bmatrix}0 & -2 & -2\\0 & -3 & 3\end{bmatrix}$
$\Rightarrow LHS=RHS.$
$\Rightarrow (A-B)'=A'-B'.$