Question
Find the value of $x$ if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$.

Answer

$\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$$
\begin{aligned}
\text { LHS } & =\left|\begin{array}{cc}
2 & 4 \\
5 & 1
\end{array}\right|=2-20=-18 \\
\text { RHS } & =\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|=\left(2 x^2-24\right) \\
\therefore \quad \text { Hence } \quad-18 & =2 x^2-24 \text { or } 2 x^2=24-18=6 \\
\therefore \quad x^2 & =3 \Rightarrow x= \pm \sqrt{3} \quad \text {}
\end{aligned}
$

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