Q115 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
If $A=\left[\begin{array}{rr}\cos \dfrac{2 \pi}{3} & -\sin \dfrac{2 \pi}{3} \\ \sin \dfrac{2 \pi}{3} & \cos \dfrac{2 \pi}{3}\end{array}\right]$ and I is an identity matrix of order $2 \times 2$ then $A^3=$
- Correct answerAI
- BNull Matrix
- C8 I
- D-2 I
Solution
The matrix $A$ is given by $A = \begin{bmatrix} \cos \dfrac{2 \pi}{3} & -\sin \dfrac{2 \pi}{3} \\ \sin \dfrac{2 \pi}{3} & \cos \dfrac{2 \pi}{3} \end{bmatrix}$.
This is a rotation matrix of the form $R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ where $\theta = \dfrac{2 \pi}{3}$.
The property of rotation matrices states that $R(\theta)^n = R(n \theta)$.
Therefore, $A^3 = R\left(3 \times \dfrac{2 \pi}{3}\right) = R(2 \pi)$.
Calculating the values for $R(2 \pi)$: $R(2 \pi) = \begin{bmatrix} \cos(2 \pi) & -\sin(2 \pi) \\ \sin(2 \pi) & \cos(2 \pi) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Since $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$, it follows that $A^3 = I$.
Answer: I
This is a rotation matrix of the form $R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ where $\theta = \dfrac{2 \pi}{3}$.
The property of rotation matrices states that $R(\theta)^n = R(n \theta)$.
Therefore, $A^3 = R\left(3 \times \dfrac{2 \pi}{3}\right) = R(2 \pi)$.
Calculating the values for $R(2 \pi)$: $R(2 \pi) = \begin{bmatrix} \cos(2 \pi) & -\sin(2 \pi) \\ \sin(2 \pi) & \cos(2 \pi) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Since $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$, it follows that $A^3 = I$.
Answer: I
Hence, the correct answer is option: 'A'.