Let P(n) = \(\begin{bmatrix}1&2\\0&1\end{bmatrix}^n=\begin{bmatrix}1&2n\\0&1\end{bmatrix}\), for all n ∈ N
Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
\(\therefore\) \(\begin{bmatrix}1&2\\0&1\end{bmatrix}^n=\begin{bmatrix}1&2n\\0&1\end{bmatrix}\), for all n ∈ N.