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Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices2 Marks
Question
Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
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Answer

Given: $\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&2-2&-2+2\\4+1&\text{x}+0&6-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&0&0\\5&\text{x}&5\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\text{x}-\text{y}+3=6$
$\Rightarrow\text{x}-\text{y}=6-3$
$\Rightarrow\text{x}-\text{y}=3\ \dots(1)$
Also,
$\text{x}=2\text{x}+\text{y}$
$\Rightarrow-\text{x}=\text{y}\ \dots(2)$
Putting the value of y in eq. (1), we get
$\text{x}-(-\text{x})=3$
$\Rightarrow2\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{2}$
Putting the value of x in eq. (2), we get
$-\Big(\frac{3}{2}\Big)=\text{y}$
$\Rightarrow\text{y}=-\frac{3}{2}$

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