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Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices3 Marks
Question
Find matrices X and Y, if $2\text{X}-\text{Y}=\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}$ and $\text{X}+2\text{Y}=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
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Answer

Given: $(2\text{X}-\text{Y})=\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}\ \dots(1)$
$(\text{X}+2\text{Y})=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}\ \dots(2)$
Multiplying eq. (1) by eq. (2), we get
$2(2\text{X}-\text{Y})=2\begin{bmatrix}6&-6&0\\-4&2&1\end{bmatrix}$
$\Rightarrow4\text{X}-2\text{Y}=\begin{bmatrix}12&-12&0\\-8&4&2\end{bmatrix}\ \dots(3)$
From eq. (3) and eq. (4), we get
$(4\text{X}-2\text{Y})+(\text{X}+2\text{Y})=\begin{bmatrix}12&-12&0\\-8&4&2\end{bmatrix}+\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
$\Rightarrow5\text{X}=\begin{bmatrix}12+3&-12+2&0+5\\-8-2&4+1&2-7\end{bmatrix}$
$\Rightarrow5\text{X}=\begin{bmatrix}15&-10&5\\-10&5&-5\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}15&-10&5\\-10&5&-5\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}$
Putting the value of X in eq. (2), we get
$(\text{X}+2\text{Y})=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}+2\text{Y}=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}$
$\Rightarrow2\text{Y}=\begin{bmatrix}3&2&5\\-2&1&-7\end{bmatrix}-\begin{bmatrix}3&-2&1\\-2&1&-1\end{bmatrix}$
$\Rightarrow2\text{Y}=\begin{bmatrix}3-3&2+2&5-1\\-2+2&1-1&-7+1\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}0&2&2\\0&0&-3\end{bmatrix}$

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