(a) To show that A is non-singular, we need to show that its determinant is non-zero for all real \(k\).

The determinant of \(A\) is calculated as follows:

\(\det(A) = k \begin{vmatrix} 5 & 2 \\ 3 & -k \end{vmatrix} - 1 \begin{vmatrix} 6 & 2 \\ -1 & -k \end{vmatrix} + 0 \begin{vmatrix} 6 & 5 \\ -1 & 3 \end{vmatrix}\)

\(= k(5(-k) - 6) - (6(-k) + 2)\)

\(= k(-5k - 6) - (-6k + 2)\)

\(= -5k^2 - 6k + 6k - 2\)

\(= -5k^2 - 2\)

Since \(-5k^2 - 2\) is always negative for all real \(k\), the determinant is never zero. Therefore, A is non-singular.

(b) Given \(A^{-1} = \begin{pmatrix} 3 & 0 & -1 \\ 1 & 0 & 0 \\ -\frac{23}{2} & \frac{1}{2} & 3 \end{pmatrix}\), we use the property \(AA^{-1} = I\).

From the first row of \(A\) and the first column of \(A^{-1}\), we have:

\(k \cdot 3 + 1 \cdot 1 + 0 \cdot -\frac{23}{2} = 1\)

\(3k + 1 = 1\)

\(3k = 0\)

\(k = 0\)