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Determinants and Matrices — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDeterminants and Matrices3 Marks
Question
Find $x$ and $y$, if $\left\{4\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 0 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -3 & 4 \\ 2 & 1 & 1\end{array}\right]\right\}\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
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Answer

$\begin{aligned} & \left\{4\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 0 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -3 & 4 \\ 2 & 1 & 1\end{array}\right]\right\}\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]\end{aligned}  $
$ \therefore\left\{\left[\begin{array}{ccc}8 & -4 & 12 \\ 4 & 0 & 8\end{array}\right]-\left[\begin{array}{ccc}3 & -3 & 4 \\ 2 & 1 & 1\end{array}\right]\right\}\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\begin{aligned} & \therefore \left[\begin{array}{ccc}8-3 & -4+3 & 12-4 \\ 4-2 & 0-1 & 8-1\end{array}\right]\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right] \end{aligned} $
$ \therefore \left[\begin{array}{lll}5 & -1 & 8 \\ 2 & -1 & 7\end{array}\right]\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\begin{aligned} & \therefore \left[\begin{array}{c}10+1+8 \\ 4+1+7\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right] \end{aligned} $
$ \therefore \left[\begin{array}{l}19 \\ 12\end{array}\right]=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\therefore$ By equality of matrices, we get $\mathrm{x}=19$ and $\mathrm{y}=12$

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