CAIE A-Level Further Math AS 1.7.1 Proof By Induction Questions

$2 \mathbf{A}=\left(\begin{array}{ll}6 2 2\end{array}\right)=\left(\begin{array}{ll}2 \times 3 3-1 2\end{array}\right)$ so true when n=1.
States base case.

Assume that it is true for n=k, so $2 \mathbf{A}^{k}=\left(\begin{array}{ll}2 \times 3^{k} 3^{k}-1 2\end{array}\right)$.
States inductive hypothesis.

Then $2 \mathbf{A}^{k+1}=\left(\begin{array}{ll}2 \times 3^{k} 3^{k}-1 2\end{array}\right)\left(\begin{array}{ll}3 1 1\end{array}\right)=\left(\begin{array}{cc}2 \times 3^{k+1} 3^{k+1}-3+2 2\end{array}\right)$
Multiplies $2 \mathbf{A}^{k}$ with A.

So, it is also true for n=k+1. Hence, by induction, true for all positive integers.
States conclusion.

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