(a) The matrix \(\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) represents a shear in the \(x\)-direction.

(b) Base case: For \(n = 1\), \(\mathbf{A}^1 = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\), which is true.

Inductive step: Assume true for \(n = k\), so \(\mathbf{A}^k = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\).

Then \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a + ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1)a \\ 0 & 1 \end{pmatrix}\).

Thus, by induction, \(\mathbf{A}^n = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix}\) for all positive integers \(n\).

(c) \(\mathbf{A}^n \mathbf{B} = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix} \begin{pmatrix} b & b \\ a^{-1} & a^{-1} \end{pmatrix} = \begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\).

Transform \(\begin{pmatrix} x \\ y \end{pmatrix}\) by multiplying matrices to find \(\begin{pmatrix} X \\ Y \end{pmatrix}\).

\(\begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (b+n)(x+y) \\ \frac{1}{a}(x+y) \end{pmatrix}\).

\(\frac{1}{a}(1+m) = m(b+n)(1+m)\)

\(y = -x\)

\(y = \frac{1}{a(b+n)}x\)