If A = 12 bmatrix 1 & 3 - 3 & 1 bmatrix, then: — Solution | JEE Mathematics

MCQMediumJEE 2023Algebra of Matrices

If $A = \frac{1}{2} \begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}$ , then:

Answer

Correct answer:C

Standard Method

Given: $A = \frac{1}{2} \begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}$

Find: Which given relation is true.

Express $A$ in trigonometric form:

A = \begin{bmatrix} \cos 60^\circ & \sin 60^\circ \\ -\sin 60^\circ & \cos 60^\circ \end{bmatrix}

Here, $\alpha = \frac{\pi}{3}$ .

For a rotation matrix,

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

Now,

A^{30} = \begin{bmatrix} \cos 30\alpha & \sin 30\alpha \\ -\sin 30\alpha & \cos 30\alpha \end{bmatrix}

Since $30\alpha = 10\pi$ , we get $\cos 30\alpha = 1$ and $\sin 30\alpha = 0$ . Therefore,

A^{30} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

Similarly,

A^{25} = \begin{bmatrix} \cos 25\alpha & \sin 25\alpha \\ -\sin 25\alpha & \cos 25\alpha \end{bmatrix}

Using periodicity, $25\alpha = \frac{25\pi}{3} = 8\pi + \frac{\pi}{3}$ , hence $A^{25} = A$ .

Now check the relation:

A^{30} + A^{25} - A = I + A - A = I

Therefore, the correct option is C.

Treating $A$ as an arbitrary matrix and trying long multiplication for high powers is inefficient. This matrix is a rotation matrix, so use its trigonometric form and periodicity instead.
Using the wrong sign pattern for a rotation matrix is incorrect. The form here is $\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ , so identify the angle carefully before computing powers.
Reducing $25\alpha$ and $30\alpha$ incorrectly modulo $2\pi$ leads to wrong conclusions. Always write $25\alpha = \frac{25\pi}{3} = 8\pi + \frac{\pi}{3}$ and $30\alpha = 10\pi$ before evaluating sine and cosine.

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