(a) To find when \(A\) is non-singular, set \(\det(A) \neq 0\). Calculate \(\det(A) = 1(k \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - k \cdot 7)\).

Simplifying, \(\det(A) = -12k + 60\). Set \(-12k + 60 \neq 0\), giving \(k \neq 5\).

(b) The inverse of \(A\) is \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\). The top row of \(\text{adj}(A)\) is \(\begin{pmatrix} k \cdot 9 - 6 \cdot 8 & -(4 \cdot 9 - 6 \cdot 7) & 4 \cdot 8 - k \cdot 7 \end{pmatrix}\).

Thus, the top row of \(A^{-1}\) is \(\begin{pmatrix} \frac{1}{60 - 12k} & \frac{6}{60 - 12k} & \frac{3k - 16}{4(5-k)} \end{pmatrix}\).

(c) Calculate \(BA = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & k & 6 \\ 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 4 & k & 6 \end{pmatrix}\).

Set \(C = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}\) such that \(BAC = \begin{pmatrix} 2 & 1 \\ k & 4 \end{pmatrix}\).

Choose \(C = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\).

(d) The transformation matrix is \(\begin{pmatrix} 2 & 1 \\ k & 4 \end{pmatrix}\). The characteristic equation is \(\lambda^2 - 6\lambda + (8-k) = 0\).

For two distinct invariant lines, the discriminant must be positive: \(6^2 - 4(8-k) > 0\).

Simplifying, \(36 - 32 + 4k > 0\) gives \(4k > -4\), so \(k > 1\). Also, \(k \neq 8\) to avoid repeated roots.