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

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

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Answer: Hence proved,

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

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