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Matrices (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)4 Marks
Question
Find $x, y, z$ if $\left\{5\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]-\left[\begin{array}{cc}2 & 1 \\ 3 & -2 \\ 1 & 3\end{array}\right]\right\}\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]$
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Answer

$\begin{aligned} & \left\{\left[\begin{array}{ll}5 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]-\left[\begin{array}{cc}2 & 1 \\ 3 & -2 \\ 1 & 3\end{array}\right]\right\}\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]\end{aligned}  $
$ \therefore\left\{\left[\begin{array}{ll}0 & 5 \\ 5 & 0 \\ 5 & 5\end{array}\right]-\left[\begin{array}{cc}2 & 1 \\ 3 & -2 \\ 1 & 3\end{array}\right]\right\}\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right]$
$\begin{aligned} & \therefore\left[\begin{array}{cc}0-2 & -1 \\ 5-3 & 0+2 \\ 5-1 & -3\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{c}x-1 \\ y+1 \\ 2 z\end{array}\right] \end{aligned} $
$ \therefore\left[\begin{array}{cr}-2 & 4 \\ 2 & 2 \\ 4 & 2\end{array}\right]\left[\begin{array}{l}2 \\ 1\end{array}\right]=\left[\begin{array}{l}x-1 \\ y+1 \\ 2 z\end{array}\right]$
$\begin{aligned} & \therefore\left[\begin{array}{c}-4+4 \\ 4+2 \\ 8+2\end{array}\right]=\left[\begin{array}{l}x-1 \\ y+1 \\ 2 z\end{array}\right] \end{aligned} $
$ \therefore\left[\begin{array}{c}0 \\ 6 \\ 10\end{array}\right]=\left[\begin{array}{l}x-1 \\ y+1 \\ 2 z\end{array}\right]$
$\therefore$ By equality of matrices, we get
$x-1=0 \therefore x=1$
$ y+1=6 \therefore y=5$
$ 2 z=10 \therefore z=5$

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