Answer
$x=1$
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:
$D=x(1\cdot x-x\cdot1)-(1)(1\cdot x-x\cdot x)+1(1\cdot 1-1\cdot x)=x(0)-1(x-x^2)+1(1-x)=x^2-x+1-x=x^2-2x+1$.
We know that this is equal to $0$, so
$x^2-2x+1=(x-1)^2=0\\(x-1)=0\\x=1$
