NIMCET 2020 — Mathematics PYQ

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Question

NIMCET 2020 — Mathematics PYQ

NIMCET | Mathematics | 2020

If $A = [cos α - sin α am p; sin α am p; cos α]$ , then for any positive integer n, $A^{n}$ is:

Choose the correct answer:

Explanation

Solution

1. Given Matrix:

A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}

2. Calculating $A^2$ :

A^2 = A \cdot A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}

A^2 = \begin{bmatrix} (\cos^2 \alpha - \sin^2 \alpha) & (\cos \alpha \sin \alpha + \sin \alpha \cos \alpha) \\ (-\sin \alpha \cos \alpha - \cos \alpha \sin \alpha) & (-\sin^2 \alpha + \cos^2 \alpha) \end{bmatrix}

3. Applying Trigonometric Identities:

Using $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ and $\sin 2\alpha = 2\sin \alpha \cos \alpha$ :

A^2 = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix}

4. Calculating $A^3$ :

A^3 = A^2 \cdot A = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}

A^3 = \begin{bmatrix} (\cos 2\alpha \cos \alpha - \sin 2\alpha \sin \alpha) & (\cos 2\alpha \sin \alpha + \sin 2\alpha \cos \alpha) \\ (-\sin 2\alpha \cos \alpha - \cos 2\alpha \sin \alpha) & (-\sin 2\alpha \sin \alpha + \cos 2\alpha \cos \alpha) \end{bmatrix}

Using $\cos(x+y)$ and $\sin(x+y)$ formulas:

A^3 = \begin{bmatrix} \cos 3\alpha & \sin 3\alpha \\ -\sin 3\alpha & \cos 3\alpha \end{bmatrix}

5. Generalization for $n$ :

By observing the pattern for $n = 1, 2, 3$ , we can conclude by Mathematical Induction:

A^n = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{bmatrix}

Final Answer:

The value of $A^n$ is:

A^n = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{bmatrix}