Q.4 Check whether the matrix \(\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\) is invertible or not.
Best answer
Answer: The matrix is invertible, since its determinant is 1 (non-zero).
Let \(A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\).
\(|A| = \cos\theta \cdot \cos\theta - (\sin\theta)(-\sin\theta) = \cos^2\theta + \sin^2\theta = 1 \ne 0\).
\(A\) is a non-singular square matrix, so \(A^{-1}\) exists for every \(\theta\). Hence \(A\) is invertible.
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