(a) The transformation matrix for a stretch in the x-direction by a factor of 3 is \(\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\). The reflection in the line y = x is represented by \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). Therefore, \(\mathbf{M} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix}\).

(b) The inverse of \(\mathbf{M}\) is \(\mathbf{M}^{-1} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix}\). This represents a reflection in the line y = x followed by a stretch in the x-direction by a factor of \(\frac{1}{3}\).

(c) Given \(\mathbf{MN} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\), we find \(\mathbf{N}\) by multiplying by \(\mathbf{M}^{-1}\) on the left: \(\mathbf{N} = \mathbf{M}^{-1} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} \\ 1 & 2 \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ 3x + 2y \end{pmatrix}\). Setting \(y = mx\), we have \(3x + 2mx = m(x + 2y)\). This leads to the quadratic \(2m^2 - m - 3 = 0\), which factors to \((2m + 3)(m - 1) = 0\). Thus, \(m = \frac{3}{2}\) or \(m = -1\), giving the lines \(y = \frac{3}{2}x\) and \(y = -x\).