Answer:

1. (a) \(d=15\).
2. (b) \(\displaystyle \mathbf A^n=\begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\) for all positive integers \(n\).
3. (c) \(n=5\).

(a) The area scale factor of a matrix transformation is the absolute value of its determinant. For \(\mathbf A=\begin{pmatrix}2&0\\1&1\end{pmatrix}\),

\(\det \mathbf A=2\), so \(\det(\mathbf A^{-1})=(\det \mathbf A)^{-1}=\frac{1}{2}\).

Therefore the image area is \(30\times \frac{1}{2}=15\), so \(d=15\).

(b) We prove by induction that

\(\displaystyle \mathbf A^n=\begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\).

For \(n=1\),

\(\displaystyle \mathbf A=\begin{pmatrix}2&0\\1&1\end{pmatrix}=\begin{pmatrix}2^1&0\\2^1-1&1\end{pmatrix}\),

so the result is true when \(n=1\).

Assume it is true for \(n=k\), where \(k\) is a positive integer. Then

\(\displaystyle \mathbf A^k=\begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}\).

Multiplying by \(\mathbf A\),

\(\displaystyle \mathbf A^{k+1}=\mathbf A^k\mathbf A=\begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}\begin{pmatrix}2&0\\1&1\end{pmatrix}=\begin{pmatrix}2^{k+1}&0\\2(2^k-1)+1&1\end{pmatrix}\).

Since \(2(2^k-1)+1=2^{k+1}-1\),

\(\displaystyle \mathbf A^{k+1}=\begin{pmatrix}2^{k+1}&0\\2^{k+1}-1&1\end{pmatrix}\).

So, by induction, the formula is true for all positive integers \(n\).

(c) Using part (b),

\(\displaystyle \mathbf A^n\mathbf B=\begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\begin{pmatrix}1&0\\33&0\end{pmatrix}=\begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}\).

A point \((x,y)\) is transformed to

\(\displaystyle \begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2^n x\\(2^n+32)x\end{pmatrix}\).

For the line \(y=2x\) to be invariant, the image must also satisfy \(y=2x\). Hence

\((2^n+32)x=2(2^n x)\).

For non-zero points on the line, this gives \(2^n+32=2^{n+1}\), so \(2^n=32\). Therefore \(n=5\).