The matrix […] is non-singular for

The correct answer is Option (4) → all real values of x except 2 and \(-2\)

A square matrix is non-singular when its determinant is not zero. Expanding along the first row:

    \( \displaystyle \Delta = -x\begin{vmatrix} x & -x \\ -2 & -x \end{vmatrix} - x\begin{vmatrix} 2 & -x \\ x & -x \end{vmatrix} + 2\begin{vmatrix} 2 & x \\ x & -2 \end{vmatrix} \)

    \(\Delta = -x(-x^2 - 2x) - x(-2x + x^2) + 2(-4 - x^2)\)

    \(\Delta = x^3 + 2x^2 + 2x^2 - x^3 - 8 - 2x^2\)

    \(\Delta = 2x^2 - 8\)

The matrix is non-singular when

    \(2x^2 - 8 \ne 0\)
    \(x^2 \ne 4\)
    \(x \ne 2,\quad x \ne -2\)

So the matrix is non-singular for all real x except 2 and −2; at those two values it is singular.