Answer
$x=\pm1$
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 is given as
$D=a(ei-fh)-b(di-fg)+c(dh-eg).$
Thus, we have:
$D=1(1\cdot 1-0\cdot0)-(0)(x^2\cdot 1-0\cdot x)+x(x^2\cdot 0-1\cdot x)=1(1)-0(x^2)+x(-x)=-x^2+1=-(x+1)(x-1)$.
We know that this is equal to $0$, so $-(x+1)(x-1)=0$
Hence,
$x=\pm1$
