Question:

If
\[ A = \begin{bmatrix} \cos\theta & -\sin\theta \\ -\sin\theta & -\cos\theta \\ \end{bmatrix} \] then \(A^{-1} =\) 

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If a matrix satisfies \(A^2 = I\), then \(A^{-1} = A\).
Updated On: Feb 2, 2026
  • \( \begin{bmatrix} -\sin\theta & -\cos\theta \\ -\cos\theta & \sin\theta \end{bmatrix} \)
  • \( \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & -\sin\theta \end{bmatrix} \)
  • \( \begin{bmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \)
  • \( \begin{bmatrix} \cos\theta & -\sin\theta \\ -\sin\theta & -\cos\theta \end{bmatrix} \)
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The Correct Option is D

Solution and Explanation

Step 1: Find the determinant of \(A\).
\[ |A| = (\cos\theta)(-\cos\theta) - (-\sin\theta)(-\sin\theta) = -\cos^2\theta - \sin^2\theta = -1 \]
Step 2: Use the inverse formula for a \(2 \times 2\) matrix.
\[ A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
Step 3: Substitute the values.
\[ A^{-1} = \begin{bmatrix} \cos\theta & -\sin\theta \\ -\sin\theta & -\cos\theta \end{bmatrix} \]
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