4 [ ccc 2 & -1 & 3 1 & 0 & 2 ]- [ ccc 3 & -3 & 4 2 & 1 & 1 ] [ c … - Vidyadip

Question

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

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Answer

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

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