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 $(\text{A}-\text{B})\text{C}=\text{AC}-\text{BC}.$
✓
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$(\text{A}-\text{B})=\begin{bmatrix}1-4&2-0\\-1-1&3-5\end{bmatrix}=\begin{bmatrix}-3&2\\-2&-2\end{bmatrix}$
$(\text{A}-\text{B})\text{C}=\begin{bmatrix}-3&2\\-2&-2\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$
$=\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}\ ....(\text{i})$
Now, $\text{AC}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$ $=\begin{bmatrix}4&-4\\1&-6\end{bmatrix}\ .....(\text{ii})$ and $\text{BC}=\begin{bmatrix}4&0\\1&5\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$ $=\begin{bmatrix}8&0\\7&-10\end{bmatrix}\ ....(\text{iii})$ $\therefore\ \text{AC}-\text{BC}=\begin{bmatrix}4-8&-4-0\\1-7&-6+10\end{bmatrix}$ [Using (ii) and (iii)] $=\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}$ $=(\text{A}-\text{B})\text{C}$ [Using (i)]Hence proved.
$(\text{A}-\text{B})\text{C}=\begin{bmatrix}-3&2\\-2&-2\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$
$=\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}\ ....(\text{i})$
Now, $\text{AC}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$ $=\begin{bmatrix}4&-4\\1&-6\end{bmatrix}\ .....(\text{ii})$ and $\text{BC}=\begin{bmatrix}4&0\\1&5\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$ $=\begin{bmatrix}8&0\\7&-10\end{bmatrix}\ ....(\text{iii})$ $\therefore\ \text{AC}-\text{BC}=\begin{bmatrix}4-8&-4-0\\1-7&-6+10\end{bmatrix}$ [Using (ii) and (iii)] $=\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}$ $=(\text{A}-\text{B})\text{C}$ [Using (i)]Hence proved.
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