Question
If \[ A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] prove that \[ A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \] for all positive integers \(n\).
Solution (Mathematical Induction)
Step 1: Base Case (n = 1)
\[ A^1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] ✔ True for \(n=1\)Step 2: Assume for \(n = k\)
\[ A^k = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \]Step 3: Prove for \(n = k+1\)
\[ A^{k+1} = A^k \cdot A \] \[ = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & k+1 \\ 0 & 1 \end{bmatrix} \]Step 4: Conclusion
Since true for \(n=1\) and true for \(k \Rightarrow k+1\), \[ A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \quad \forall n \in \mathbb{N} \]Final Result
\[
A^n =
\begin{bmatrix}
1 & n \\
0 & 1
\end{bmatrix}
\]
Hence proved.