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If \( 2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right] \) then the value of \( x \) and \( y \) are
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Verified Answer
The correct answer is:
\( x=3, y=3 \)
Given that,
$2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
$\left[\begin{array}{cc}2+y & 6 \\ 1 & 2 x+2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
Equating first elements, we have
$2+y=5 \Rightarrow y=3$
Equating last elements, we have
$2 x+2=8 \Rightarrow x=3$
Therefore, $x=3, y=3$.
$2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
$\left[\begin{array}{cc}2+y & 6 \\ 1 & 2 x+2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
Equating first elements, we have
$2+y=5 \Rightarrow y=3$
Equating last elements, we have
$2 x+2=8 \Rightarrow x=3$
Therefore, $x=3, y=3$.
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