3 [ ll 2 & 0 0 & 2 2 & 2 ]-4 [ cc 1 & 1 -1 & 2 3 & 1 ] [ l 1 2 ]=… - Vidyadip

Question

$\left\{3\left[\begin{array}{ll}2 & 0 \\ 0 & 2 \\ 2 & 2\end{array}\right]-4\left[\begin{array}{cc}1 & 1 \\ -1 & 2 \\ 3 & 1\end{array}\right]\right\}\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$

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Answer

$\therefore\left\{\left[\begin{array}{ll}6 & 0 \\ 0 & 6 \\ 6 & 6\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ -4 & 8 \\ 12 & 4\end{array}\right]\right\}\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore\left[\begin{array}{ll}6-4 & 0-4 \\ 0+4 & 6-8 \\ 6-12 & 6-4\end{array}\right]\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore\left[\begin{array}{rr}2 & -4 \\ 4 & -2 \\ -6 & 2\end{array}\right]\left[\begin{array}{l}1 \\ 2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore\left[\begin{array}{c}2-8 \\ 4-4 \\ -6+4\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore \left[\begin{array}{c}-6 \\ 0 \\ -2\end{array}\right]=\left[\begin{array}{c}x-3 \\ y-1 \\ 2 z\end{array}\right]$
$\therefore$ By equality of matrices,
we get $x-3=-6, y-1=0,2 z=-2 $
$\therefore x=-3, y=1, z=-1$

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