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Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices2 Marks
Question
Find matrices X and Y, if $\text{X}+\text{Y}=\begin{bmatrix}5&2\\0&9\end{bmatrix}$ and $\text{X}-\text{Y}=\begin{bmatrix}3&6\\0&-1\end{bmatrix}$
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Answer

Given: $(\text{X}+\text{Y})+(\text{X}-\text{Y})=\begin{bmatrix}5&2\\0&9\end{bmatrix}+\begin{bmatrix}3&6\\0&-1\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}5+3&2+6\\0+0&9-1\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}8&8\\0&8\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}8&8\\0&8\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}4&4\\0&4\end{bmatrix}$
Now,
$(\text{X}+\text{Y})-(\text{X}-\text{Y})=\begin{bmatrix}5&2\\0&9\end{bmatrix}-\begin{bmatrix}3&6\\0&-1\end{bmatrix}$
$\Rightarrow\text{X}+\text{Y}-\text{X}+\text{Y}=\begin{bmatrix}5-3&2-6\\0-0&9+1\end{bmatrix}$
$\Rightarrow2\text{Y}=\begin{bmatrix}2&-4\\0&10\end{bmatrix}$
$\Rightarrow\text{Y}=\frac{1}{2}\begin{bmatrix}2&-4\\0&10\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}1&-2\\0&5\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}4&4\\0&4\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}1&-2\\0&5\end{bmatrix}$

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