📘 Question
If
\[
A =
\begin{bmatrix}
1 & a \\
0 & 1
\end{bmatrix}
\]
Find \(A^n\), where \(n \in \mathbb{N}\).
✏️ Step-by-Step Solution
Step 1: Observe pattern
Compute first few powers:
\[
A^2 =
\begin{bmatrix}
1 & 2a \\
0 & 1
\end{bmatrix}
\]
\[
A^3 =
\begin{bmatrix}
1 & 3a \\
0 & 1
\end{bmatrix}
\]
Step 2: Generalize
Pattern suggests:
\[
A^n =
\begin{bmatrix}
1 & na \\
0 & 1
\end{bmatrix}
\]
Step 3: Reason
Each multiplication adds another \(a\) in the upper-right position.
✅ Final Answer
\[
\boxed{
\begin{bmatrix}
1 & na \\
0 & 1
\end{bmatrix}
}
\]
💡 Key Concept
This is a special upper triangular matrix where powers increase linearly in the off-diagonal term.