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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 Marks
Question
If $\text{A}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\text{B}=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix},$ verify that A(B - C) = AB - AC.
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Answer

Given, $\text{A}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\text{B}=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ $\text{C}=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ $ \text{A}(\text{B}-\text{C})=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}\end{bmatrix}$ $=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}0-1&5-5&-4-2\\-2+1&1-1&3-0\\-1-0&0+1&2-1\end{bmatrix}$ $=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}-1&0&-6\\-1&0&3\\-1&1&1\end{bmatrix}$ $=\begin{bmatrix}-1+0+2&0+0-2&-6+0-2\\-3+1+0&0+0+0&-18-3+0\\2-1-1&0+0+1&12+3+1\end{bmatrix}$ $\text{A}(\text{B}-\text{C})=\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}\ \dots(\text{i})$ $ \text{AB}-\text{AC}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ $=\begin{bmatrix}0+0+2&5+0+0&-4+0-4\\0+2+0&15-1+0&-12-3+0\\0-2-1&-10+1+0&8+3+2\end{bmatrix}-\begin{bmatrix}1+0+0&5+0+2&2+0-2\\3+1+0&15-1+0&6+0+0\\0-2-1&-10+1+1&-4+0+1\end{bmatrix}$ $=\begin{bmatrix}2&5&-8\\2&14&-15\\-3&-9&13\end{bmatrix}-\begin{bmatrix}1&7&0\\4&14&6\\-3&-10&-3\end{bmatrix}$ $=\begin{bmatrix}2-1&5-7&-8-0\\2-4&14-14&-14-6\\-3+3&-9+10&3+3\end{bmatrix}$ $\text{AB}-\text{AC}=\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}\ \dots(\text{ii})$ From equation (i) and (ii),$\text{A}(\text{B}-\text{C})=\text{AB}-\text{AC}$

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