Answer: Let \(P_n\) be the statement that \(\mathbf{A}^n=\left(\begin{array}{cc}2^n & 3(2^n-1)\\0 & 1\end{array}\right)\).

For \(n=1\),

\(\mathbf{A}^1=\left(\begin{array}{ll}2 & 3\\0 & 1\end{array}\right)=\left(\begin{array}{cc}2^1 & 3(2^1-1)\\0 & 1\end{array}\right)\),

so \(P_1\) is true.

Now assume that \(P_k\) is true for some positive integer \(k\), so that

\(\mathbf{A}^k=\left(\begin{array}{cc}2^k & 3(2^k-1)\\0 & 1\end{array}\right)\).

Then

\(\mathbf{A}^{k+1}=\mathbf{A}\mathbf{A}^k=\left(\begin{array}{ll}2 & 3\\0 & 1\end{array}\right)\left(\begin{array}{cc}2^k & 3(2^k-1)\\0 & 1\end{array}\right)\)

\(=\left(\begin{array}{cc}2\cdot 2^k+3\cdot 0 & 2\cdot 3(2^k-1)+3\cdot 1\\0\cdot 2^k+1\cdot 0 & 0\cdot 3(2^k-1)+1\cdot 1\end{array}\right)\)

\(=\left(\begin{array}{cc}2^{k+1} & 6(2^k-1)+3\\0 & 1\end{array}\right)=\left(\begin{array}{cc}2^{k+1} & 3(2^{k+1}-2)+3\\0 & 1\end{array}\right)\)

\(=\left(\begin{array}{cc}2^{k+1} & 3(2^{k+1}-1)\\0 & 1\end{array}\right).\)

Thus \(P_{k+1}\) is true whenever \(P_k\) is true. Since \(P_1\) is true and \(P_k\Rightarrow P_{k+1}\), it follows by mathematical induction that for every positive integer \(n\),

\(\mathbf{A}^n=\left(\begin{array}{cc}2^n & 3(2^n-1)\\0 & 1\end{array}\right).\)

Let \(P_n\) be the proposition that \(\mathbf{A}^n=\left(\begin{array}{cc}2^n & 3(2^n-1)\\0 & 1\end{array}\right)\), where \(\mathbf{A}=\left(\begin{array}{ll}2 & 3\\0 & 1\end{array}\right)\).

First we check the base case \(n=1\):

\(\mathbf{A}^1=\left(\begin{array}{ll}2 & 3\\0 & 1\end{array}\right)=\left(\begin{array}{cc}2^1 & 3(2^1-1)\\0 & 1\end{array}\right).\)

So \(P_1\) is true.

Now assume \(P_k\) is true for some positive integer \(k\). Then

\(\mathbf{A}^k=\left(\begin{array}{cc}2^k & 3(2^k-1)\\0 & 1\end{array}\right).\)

We must prove \(P_{k+1}\). Multiply by \(\mathbf{A}\):

\(\mathbf{A}^{k+1}=\mathbf{A}\mathbf{A}^k=\left(\begin{array}{ll}2 & 3\\0 & 1\end{array}\right)\left(\begin{array}{cc}2^k & 3(2^k-1)\\0 & 1\end{array}\right).\)

Carrying out the matrix multiplication gives

\(\mathbf{A}^{k+1}=\left(\begin{array}{cc}2\cdot 2^k+3\cdot 0 & 2\cdot 3(2^k-1)+3\cdot 1\\0\cdot 2^k+1\cdot 0 & 0\cdot 3(2^k-1)+1\cdot 1\end{array}\right).\)

Hence

\(\mathbf{A}^{k+1}=\left(\begin{array}{cc}2^{k+1} & 6(2^k-1)+3\\0 & 1\end{array}\right).\)

Now simplify the top-right entry:

\(6(2^k-1)+3=6\cdot 2^k-6+3=6\cdot 2^k-3=3(2^{k+1}-1).\)

Therefore

\(\mathbf{A}^{k+1}=\left(\begin{array}{cc}2^{k+1} & 3(2^{k+1}-1)\\0 & 1\end{array}\right),\)

which is exactly \(P_{k+1}\).

Since \(P_1\) is true and \(P_k\Rightarrow P_{k+1}\), it follows by mathematical induction that for every positive integer \(n\),

\(\mathbf{A}^{n}=\left(\begin{array}{cc}2^{n} & 3(2^{n}-1)\\0 & 1\end{array}\right).\)