## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$x=0$ or, $x=-\dfrac{1}{2}$
The determinant for a 3 by 3 matrix can be computed as: $Determinant (D)= \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$ We have: $D= \begin{vmatrix} x & 1 & 2 \\ 1 & x & 3 \\ 0 & 1 & 2 \\ \end{vmatrix}$ Therefore, $D=x (2x-3)-1(2-0)+2(1-0) \\=2x^2-3x$ Since, $det=-4x$ So, $2x^2-3x=-4x \implies 2x^2-3x+4x=0$ $\implies x(2x+1)=0$ $\implies x=0$ or, $x=-\dfrac{1}{2}$