Answer
$x=1$ or $x=-1$
Work Step by Step
2 by 2 determinant: $D=\left|\begin{array}{ll}
a & b\\
c & d
\end{array}\right|=ad-bc$
---
$\left|\begin{array}{ll}{x}&{1}\\{3}&{x}\end{array}\right|=x^{2}-3$,
so we are solving
$x^{2}-3=-2$
$x^{2}=1$
$x=\pm 1$
