Question

If \(A = [\begin{array}{cc}cosθ & -sinθ \\ sinθ & cosθ\end{array}]\), then the matrix \(A^{-3}\) when \(θ = \frac{π}{6}\) is equal to...

• A. \([\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}]\)
• B. \([\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}]\)
• C. \([\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}]\)
• D. \([\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}]\)

Hint:

A is a rotation matrix, so A^n rotates by n theta.

Answer 1

The Correct Option is A

Step 1: Direct computation:
$A^{-1}=A^T=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}$, because $\det A=1$.

Step 2: At theta = 30 degrees:
$A^3$ is the rotation by $90^\circ$, which is $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

Step 3: Invert:
The inverse of this matrix is its transpose: $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Option (A).

Final Answer:
A^3 is rotation by 90 degrees and its inverse is the transpose. \[ \boxed{A} \]