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 $\text{a}(\text{C}-\text{A})=\text{aC}-\text{aA}.$

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$(\text{C}-\text{A})=\begin{bmatrix}2-1&0-2\\1+1&-2-3\end{bmatrix}=\begin{bmatrix}1&-2\\2&-5\end{bmatrix}$
and $\text{a}(\text{C}-\text{A})=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}\ [\because\ \text{a}=4]$ Also, $\text{aC}-\text{aA}=\begin{bmatrix}8&0\\4&-8\end{bmatrix}-\begin{bmatrix}4&8\\-4&12\end{bmatrix}$$=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}$
$=\text{a}(\text{C}-\text{A})$
Hence proved.

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