Q63 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If $A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$, then the matrix $A^{-3}$ when $\theta = \dfrac{\pi}{6}$ is equal to...
- Correct answerA$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
- B$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
- C$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
- D$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Solution
Given $A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
Using the property of rotation matrices, $A^n = \begin{bmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{bmatrix}$ for any integer $n$.
For $n = -3$, we have:
$A^{-3} = \begin{bmatrix} \cos(-3\theta) & -\sin(-3\theta) \\ \sin(-3\theta) & \cos(-3\theta) \end{bmatrix}$
Since $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$, this simplifies to:
$A^{-3} = \begin{bmatrix} \cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{bmatrix}$
Substituting $\theta = \dfrac{\pi}{6}$, we get $3\theta = \dfrac{\pi}{2}$:
$A^{-3} = \begin{bmatrix} \cos\dfrac{\pi}{2} & \sin\dfrac{\pi}{2} \\ -\sin\dfrac{\pi}{2} & \cos\dfrac{\pi}{2} \end{bmatrix}$
$A^{-3} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
Answer: $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
Using the property of rotation matrices, $A^n = \begin{bmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{bmatrix}$ for any integer $n$.
For $n = -3$, we have:
$A^{-3} = \begin{bmatrix} \cos(-3\theta) & -\sin(-3\theta) \\ \sin(-3\theta) & \cos(-3\theta) \end{bmatrix}$
Since $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$, this simplifies to:
$A^{-3} = \begin{bmatrix} \cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{bmatrix}$
Substituting $\theta = \dfrac{\pi}{6}$, we get $3\theta = \dfrac{\pi}{2}$:
$A^{-3} = \begin{bmatrix} \cos\dfrac{\pi}{2} & \sin\dfrac{\pi}{2} \\ -\sin\dfrac{\pi}{2} & \cos\dfrac{\pi}{2} \end{bmatrix}$
$A^{-3} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
Answer: $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
Hence, the correct answer is option: 'A'.