Answer
$x=0$ or $x=-\frac{1}{2}$.
Work Step by Step
We know that for a matrix
\[
\left[\begin{array}{rrr}
a & b & c \\
d &e & f \\
g &h & i \\
\end{array} \right]
\]
the determinant, $D=a(ei-fh)-b(di-fg)+c(dh-eg).$
Hence here $D=x(x\cdot2-3\cdot1)-1(1\cdot2-3\cdot0)+2(1\cdot1-x\cdot0)=-4x\\x(2x-3)-1(2)+2(1)=-4x\\2x^2-3x+4x-2+2=0\\2x^2+x=0\\x(2x+1)=0.$
Hence by the zero product property $x=0$ or $2x+1=0\\x=-\frac{1}{2}$.
