Find values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$

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Solution We have $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$

i.e.    $\quad 3-x^{2}=3-8$

i.e.    $x^{2}=8$

Hence    $x=\pm 2 \sqrt{2}$

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