Question
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{a}+\text{b})\text{B}=\text{aB}+\text{bB}.$

Answer

We have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$(\text{a}+\text{b})\text{B}=\begin{bmatrix}4&-2\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$ $[\because$ a = 4, b = -2$]$
$=\begin{bmatrix}8&0\\2&10\end{bmatrix}$
and $\text{aB}+\text{bB}=4\text{B}-2\text{B}$ $=\begin{bmatrix}16&0\\4&20\end{bmatrix}-\begin{bmatrix}8&0\\2&10\end{bmatrix}$$=\begin{bmatrix}8&0\\2&10\end{bmatrix}$
$=(\text{a}+\text{b})\text{B}$
Hence proved.

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