Answer
$\det (A)=0$ for $x \in \{0,2,-1\}$
Work Step by Step
$\det (A)=\begin{bmatrix} 1 & -1 & x \\
2& 1 & x^2\\
4 & -1 & x^3 \end{bmatrix} \approx^1\begin{bmatrix}
1 & -1 & x \\
0& 3 & x^2-2x\\
0 &3 & x^3-4x \end{bmatrix} \approx^2 \begin{bmatrix}
1 & -1 & x \\
0& 3 & x^2-2x\\
0 &3 & x^3-x^2-2x \end{bmatrix} $
Also $\begin{bmatrix}
1 & -1 & x \\
0& 3 & x^2-2x\\
0 &3 & x^3-x^2-2x \end{bmatrix}=\begin{bmatrix}
1 & -1 & x \\
0& 3 & x(x-2)\\
0 &3 & x(x-2)(x+1) \end{bmatrix}$
$1. A_{12}(-2),A_{13}(-4)$
$2.A_{23}(-1)$
Hence, the determinant of the given matrix is $\det (A)=0$ for $x \in \{0,2,-1\}$
