If A=≤ft[{array}{cc} θ & - θ \\ θ & θ{array}], then the matrix A^{-3} when θ=π/6 is equal to...

Correct Answer is : [01−10]≤ft[{array}{cc}0 & 1 \\ -1 & 0{array}][0−110]

≤ft[{array}{cc}0 & 1 \\ -1 & 0{array}]

≤ft[{array}{ll}0 & 1 \\ 1 & 0{array}]

≤ft[{array}{cc}0 & -1 \\ 1 & 0{array}]

≤ft[{array}{ll}1 & 0 \\ 0 & 1{array}]

Correct Answer is : [01−10]≤ft[{array}{cc}0 & 1 \\ -1 & 0{array}][0−110]

Test Index

MHTCET 15 Apr 2026 Shift 1 Paper

Question : 116 of 150

Marks: +1, -0

A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]

A^{-3}

\theta=\frac{\pi}{6}

is equal to...

Solution:

Concept:

A rotation matrix raised to any integer power simply scales the angle by that power.

Explanation:

The given matrix is a rotation matrix.

For a rotation matrix,

A^{n}=\begin{bmatrix}\cos(n\theta) & -\sin(n\theta)\\ \sin(n\theta) & \cos(n\theta)\end{bmatrix}

Therefore,

A^{-3}=\begin{bmatrix}\cos(-3\theta) & -\sin(-3\theta)\\ \sin(-3\theta) & \cos(-3\theta)\end{bmatrix}

Given

\theta=\frac{\pi}{6}

-3\theta=-3\cdot\frac{\pi}{6}=-\frac{\pi}{2}

Now,

\cos\left(-\frac{\pi}{2}\right)=0

and

\sin\left(-\frac{\pi}{2}\right)=-1

Substituting these values gives,

A^{-3}=\begin{bmatrix}0 & -(-1)\\ -1 & 0\end{bmatrix}=\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}

This matches option A.

Answer:

Option A:

\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}

Go to Question:

Prev Question

Next Question