#### Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question 59

$A^{n}=\left[\begin{array}{cc}\cos n \alpha+\sin n \propto & \sqrt{2} \sin n \propto \\ -\sqrt{2} \sin n \propto & \cos n \alpha-\sin n \alpha\end{array}\right]$ for all $n \in N$

Hint: We use the principle of mathematical induction.

Given:

$A=\left[\begin{array}{cc}\cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \propto & \cos \alpha-\sin \alpha\end{array}\right]$

Prove:

$A^{n}=\left[\begin{array}{cc}\cos n \alpha+\sin n \propto & \sqrt{2} \sin n \propto \\ -\sqrt{2} \sin n \propto & \cos n \alpha-\sin n \alpha\end{array}\right]$  for all $n \in N$

Solution:

Step 1: Put n=1 in  eqn (i)

$A^{1}=\left[\begin{array}{cc}\cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \alpha & \cos \alpha-\sin \alpha\end{array}\right]$

$A^n$  is true for n=1

Step 2: Let, $A^n$  is true for n=k

So, $A^{k}=\left[\begin{array}{cc}\cos k \propto+\sin k \propto & \sqrt{2} \sin k \propto \\ -\sqrt{2} \sin k \propto & \cos k \propto-\sin k \alpha\end{array}\right]$

Step 3: Now, we have to show that $A^n$ is true for n=k+1

$A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \propto+\sin (k+1) \propto & \sqrt{2} \sin (k+1) \propto \\ -\sqrt{2} \sin (k+1) \propto & \cos (k+1) \propto-\sin (k+1) \propto\end{array}\right]$

Now, $A^{k+1}=A^{k} A$

$\\=\left[\begin{array}{cc} \cos k \alpha+\sin k \propto & \sqrt{2} \sin k \propto \\ -\sqrt{2} \sin k \propto & \cos k \propto-\sin k \propto \end{array}\right]\left[\begin{array}{cc} \cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \propto & \cos \alpha-\sin \alpha \end{array}\right]\\\\\\ =\left[\begin{array}{cc} (\cos k \propto+\sin k \propto)(\cos \alpha+\sin \propto) & (\cos k \propto+\sin k \propto) \sqrt{2} \sin \propto \\ -2 \sin \alpha \sin k \propto & +\sqrt{2} \sin k \propto(\cos \alpha-\sin \alpha) \\ (\cos \alpha+\sin \propto)(-\sqrt{2} \sin k \propto) & -2 \sin k \propto \sin \propto \\ -\sqrt{2} \sin \alpha(\cos k \propto-\sin k \propto) & +(\cos k \propto-\sin k \propto)(\cos \alpha-\sin \alpha) \end{array}\right]\\$

$=\left[\begin{array}{cc} \cos k \propto \cos \alpha+\sin k \propto \cos \alpha+\cos k \propto \sin \propto & \sqrt{2} \cos k \propto \sin \propto+\sqrt{2} \sin \propto \sin k \propto \\ +\sin \alpha \sin k \propto-2 \sin \propto \sin k \propto & +\sqrt{2} \sin k \propto \cos \propto-\sqrt{2} \sin k \propto \sin \propto \\ -\sqrt{2} \cos \propto \sin k \propto-\sqrt{2} \sin \propto \sin k \propto & -2 \sin k \propto \sin \propto+\cos k \propto \cos \alpha-\cos \alpha \sin k \propto \\ -\sqrt{2} \sin \propto \cos k \propto+\sqrt{2} \sin \propto \sin k \propto & -\sin \propto \cos k \propto+\sin \alpha \sin k \propto \end{array}\right]\\$

$=\left[\begin{array}{cc} \cos \alpha \cos k \propto-\sin \alpha \sin k \propto & \sqrt{2}(\sin k \propto \cos \propto+\cos k \propto \sin \alpha) \\ +\sin \alpha \cos k \propto+\sin k \propto \cos \alpha & \cos k \propto \cos \alpha-\sin k \propto \sin \alpha \\ -\sqrt{2}(\sin k \propto \cos \alpha+\cos k \propto \sin \alpha) & \left.\begin{array}{c} (- \sin k \propto \cos \alpha+\sin \propto \cos k \propto \end{array}\right) \end{array}\right]$

$=\left[\begin{array}{cc} \cos (k+1) \propto+\sin (k+1) \propto & \sqrt{2} \sin (k+1) \propto \\ -\sqrt{2} \sin (k+1) \propto & \cos (k+1) \propto-\sin (k+1) \propto \end{array}\right]$

So, $A^n$ is true for n=k+1 whenever it is true for n=k

Hence, by principle of mathematical induction, $A^n$ is true for $n \in N$