JEE 2023 — Mathematics PYQ

Solution

Step 1: Identify the form of matrix $A$.

The given matrix $A$ can be written in the form of a rotation matrix:

Here, $\cos \theta =\frac{1}{2}$ and $\sin \theta =\frac{\sqrt{3}}{2}$.

This corresponds to $\theta =60^{\circ }$ or $\theta =\frac{\pi }{3}$.

Step 2: Use the property of rotation matrices.

For a rotation matrix $A_{\theta }$, we have $A_{\theta }^n=A_{n\theta }$.

Therefore:

Step 3: Calculate $A^{30}$ and $A^{25}$.

• For $n=30$:

  $$30\theta =30\times 60^{\circ }=1800^{\circ }$$

  Since $1800^{\circ }=5\times 360^{\circ }$, this is equivalent to $0^{\circ }$.

  $$A^{30}=\left[\begin{array}{cc}\cos (0^{\circ })& amp;\sin (0^{\circ })\\ -\sin (0^{\circ })& amp;\cos (0^{\circ })\end{array}\right]=\left[\begin{array}{cc}1& amp;0\\ 0& amp;1\end{array}\right]=I$$

• For $n=25$:

  $$25\theta =25\times 60^{\circ }=1500^{\circ }$$

  $1500^{\circ }=4\times 360^{\circ }+60^{\circ }$, which is equivalent to $60^{\circ }$ (or $\theta$).

  $$A^{25}=A_{\theta }=A$$

Step 4: Verify the options.

Substitute the results into the options:

• $A^{30}+A^{25}-A=I+A-A=I$

Correct Option: (3)