If A = \begin{bmatrix} \sin 2\theta & 2 \sin^{2}\theta-1 & 0 \\ \cos 2\theta & 2 \sin \theta \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} then which of the following statements is/are correct? 1. A^{-1}=adjA 2. A is skew-symmetric matrix 3. A^{-1}=A^{T} Select the correct answer using the code given below:

Explanation

Simplify elements: 2 \sin^2 \theta - 1 = -\cos 2\theta and 2 \sin \theta \cos \theta = \sin 2\theta. So, A = \begin{bmatrix} \sin 2\theta & -\cos 2\theta & 0 \\ \cos 2\theta & \sin 2\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}. Check orthogonality: A A^T = I, so A is orthogonal and A^{-1} = A^T (Statement 3 is true). Determinant |A| = \sin^2 2\theta + \cos^2 2\theta = 1. Since A^{-1} = \frac{adjA}{|A|}, we get A^{-1} = adjA (Statement 1 is true). It is clearly not skew-symmetric.

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