Question
If \[ A = \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \] prove that \[ A^n = \begin{bmatrix} a^n & b\frac{a^n – 1}{a – 1} \\ 0 & 1 \end{bmatrix} \] for every positive integer \(n\).
Solution (Mathematical Induction)
Step 1: Base Case (n = 1)
\[ A^1 = \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \] RHS: \[ \begin{bmatrix} a^1 & b\frac{a^1 – 1}{a – 1} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & b \end{bmatrix} \] ✔ True for \(n=1\)Step 2: Assume for \(n = k\)
\[ A^k = \begin{bmatrix} a^k & b\frac{a^k – 1}{a – 1} \\ 0 & 1 \end{bmatrix} \]Step 3: Prove for \(n = k+1\)
\[ A^{k+1} = A^k \cdot A \] \[ = \begin{bmatrix} a^k & b\frac{a^k – 1}{a – 1} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} a^{k+1} & a^k b + b\frac{a^k – 1}{a – 1} \\ 0 & 1 \end{bmatrix} \]Step 4: Simplify
\[ a^k b + b\frac{a^k – 1}{a – 1} = b\left[a^k + \frac{a^k – 1}{a – 1}\right] \] \[ = b\frac{a^{k+1} – 1}{a – 1} \]Step 5: Conclusion
\[ A^{k+1} = \begin{bmatrix} a^{k+1} & b\frac{a^{k+1} – 1}{a – 1} \\ 0 & 1 \end{bmatrix} \] ✔ True for \(k+1\)Final Result
\[
A^n =
\begin{bmatrix}
a^n & b\frac{a^n – 1}{a – 1} \\
0 & 1
\end{bmatrix}
\]
Hence proved.
prove A power n upper triangular matrix formula