Question
If $\text{A}=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix},$ then verify that $(\text{A}-\text{B})'=\text{A}'-\text{B}'.$

Answer

We have, $\text{A}=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix}$$(\text{A}-\text{B})=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}-\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix}$
$=\begin{bmatrix}0&0\\-2&-3\\-2&3\end{bmatrix}$
and $(\text{A}-\text{B})'=\begin{bmatrix}0&-2&-2\\0&-3&3\end{bmatrix}$
Also, $\text{A}'-\text{B}'=\begin{bmatrix}1&4&5\\2&1&6\end{bmatrix}-\begin{bmatrix}1&6&7\\2&4&3\end{bmatrix}$
$=\begin{bmatrix}0&-2&-2\\0&-3&3\end{bmatrix}$
$=(\text{A}-\text{B})'$
Hence proved.

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