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If $A=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]$ and $B=\left[\begin{array}{c}y \\ x \\ 1\end{array}\right]$ be such that $\mathrm{AB}=\left[\begin{array}{l}6 \\ 8\end{array}\right]$, then:
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Verified Answer
The correct answer is:
$y=2 x$
$y=2 x$
Let $\mathrm{A}=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]$
$$
\begin{aligned}
&\mathrm{AB}=\left[\begin{array}{ccc}
1 & 2 & x \\
3 & -1 & 2
\end{array}\right]\left[\begin{array}{l}
y \\
x \\
1
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+2 x+x \\
3 y-x+2
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+3 x \\
3 y-x+2
\end{array}\right] \\
&\Rightarrow y+3 x=6 \text { and } 3 y-x=6
\end{aligned}
$$
On solving, we get
$$
\begin{aligned}
&x=\frac{6}{5} \text { and } y=\frac{12}{5} \\
&\Rightarrow y=2 x
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{AB}=\left[\begin{array}{ccc}
1 & 2 & x \\
3 & -1 & 2
\end{array}\right]\left[\begin{array}{l}
y \\
x \\
1
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+2 x+x \\
3 y-x+2
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{l}
6 \\
8
\end{array}\right]=\left[\begin{array}{l}
y+3 x \\
3 y-x+2
\end{array}\right] \\
&\Rightarrow y+3 x=6 \text { and } 3 y-x=6
\end{aligned}
$$
On solving, we get
$$
\begin{aligned}
&x=\frac{6}{5} \text { and } y=\frac{12}{5} \\
&\Rightarrow y=2 x
\end{aligned}
$$
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