Question
If \[ A = \begin{bmatrix} \cos \alpha + \sin \alpha & \sqrt{2}\sin \alpha \\ -\sqrt{2}\sin \alpha & \cos \alpha – \sin \alpha \end{bmatrix} \] prove that \[ A^n = \begin{bmatrix} \cos n\alpha + \sin n\alpha & \sqrt{2}\sin n\alpha \\ -\sqrt{2}\sin n\alpha & \cos n\alpha – \sin n\alpha \end{bmatrix} \quad \forall n \in \mathbb{N}. \]
Solution (Mathematical Induction)
Step 1: Base Case (n = 1)
\[ A^1 = A \] ✔ Matches RHS for \(n=1\)Step 2: Assume for \(n = k\)
\[ A^k = \begin{bmatrix} \cos k\alpha + \sin k\alpha & \sqrt{2}\sin k\alpha \\ -\sqrt{2}\sin k\alpha & \cos k\alpha – \sin k\alpha \end{bmatrix} \]Step 3: Prove for \(n = k+1\)
\[ A^{k+1} = A^k \cdot A \]Step 4: Multiply
After multiplication, \[ = \begin{bmatrix} \cos k\alpha \cos \alpha – \sin k\alpha \sin \alpha + (\sin k\alpha \cos \alpha + \cos k\alpha \sin \alpha) & \sqrt{2}(\sin k\alpha \cos \alpha + \cos k\alpha \sin \alpha) \\ -\sqrt{2}(\sin k\alpha \cos \alpha + \cos k\alpha \sin \alpha) & \cos k\alpha \cos \alpha – \sin k\alpha \sin \alpha – (\sin k\alpha \cos \alpha + \cos k\alpha \sin \alpha) \end{bmatrix} \]Step 5: Use Identities
\[ \cos(k+1)\alpha = \cos k\alpha \cos \alpha – \sin k\alpha \sin \alpha \] \[ \sin(k+1)\alpha = \sin k\alpha \cos \alpha + \cos k\alpha \sin \alpha \]Step 6: Substitute
\[ A^{k+1} = \begin{bmatrix} \cos (k+1)\alpha + \sin (k+1)\alpha & \sqrt{2}\sin (k+1)\alpha \\ -\sqrt{2}\sin (k+1)\alpha & \cos (k+1)\alpha – \sin (k+1)\alpha \end{bmatrix} \]Step 7: Conclusion
✔ True for \(k+1\) \[ \Rightarrow A^n = \begin{bmatrix} \cos n\alpha + \sin n\alpha & \sqrt{2}\sin n\alpha \\ -\sqrt{2}\sin n\alpha & \cos n\alpha – \sin n\alpha \end{bmatrix} \]Final Result
\[
A^n =
\begin{bmatrix}
\cos n\alpha + \sin n\alpha & \sqrt{2}\sin n\alpha \\
-\sqrt{2}\sin n\alpha & \cos n\alpha – \sin n\alpha
\end{bmatrix}
\]
Hence proved.