Area Under the Curve
Matrix Powers — Rotation Matrix [Misplaced in PDF]
nta_pyq_2023_jan
Grade None
Question:
If $A=\dfrac{1}{2}\begin{bmatrix}1 & \sqrt{3}\\-\sqrt{3} & 1\end{bmatrix}$, then:
A^30 - A^25 = 2I
A^30 + A^25 + A = I
A^30 + A^25 - A = I
A^30 = A^25
Step-by-Step Solution
Key Concept: $A=\begin{bmatrix}\cos60^\circ & \sin60^\circ\\-\sin60^\circ & \cos60^\circ\end{bmatrix}$, a rotation by $\alpha=\pi/3$. $A^n$ rotates by $n\alpha$. $A^{30}$ rotates by $10\pi=0\Rightarrow A^{30}=I$. $A^{25}$ rotates by $25\pi/3$.
$A^{30}=I$, $A^{25}=A$. $A^{30}+A^{25}-A=I$.
Correct Answer: 3