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If $A\, = \,\left[ {\begin{array}{*{20}{c}}
1&2&x\\
3&{ - 1}&2
\end{array}} \right]$ and $B\, = \,\left[ {\begin{array}{*{20}{c}}
y\\
x\\
1
\end{array}} \right]$ be such that $AB\, = \,\left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right],$ then
$y = 2x$
$y = -2x$
$y = x$
$y = -x$
Solution
Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&x\\
3&{ – 1}&2
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
y\\
x\\
1
\end{array}} \right]$
$AB = \left[ {\begin{array}{*{20}{c}}
1&2&x\\
3&{ – 1}&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
y\\
x\\
1
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{y + 2x + x}\\
{3y – x + 2}
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{y + 3x}\\
{3y – x + 2}
\end{array}} \right]$
$ \Rightarrow y + 3x = 6$ and $3y – x = 6$
On solving, we gat
$x = \frac{6}{5}$ and $y = \frac{{12}}{5}$
$ \Rightarrow y = 2x$