Chapter 10 - Systems of Equations and Inequalities - Section 10.3 Systems of Linear Equations: Determinants - 10.3 Assess Your Understanding - Page 761: 52

$x =\pm 1$

The determinant for a 2 by 2 matrix can be computed as: $Determinant (D)= \begin{vmatrix} A & B \\ C & D\end{vmatrix}=AD-BC$ Therefore, $Det=\begin{vmatrix} x & 1 \\ 3 & x \end{vmatrix}=x^2-3$ Since $det=-2$ So, $x^2-3=-2 \implies x^2=1 $ or, $x =\pm 1$