A
To solve the question, we need to find the expression for the inverse of the adjugate of a matrix \(A\). The matrix \(A\) given in the problem is:
| \(A = \begin{vmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{vmatrix}\) |
Firstly, recall the property of the inverse of the adjugate of a matrix. For a non-singular matrix \(A\):
- The adjugate of \(A\), denoted as \(\text{adj } A\), and the inverse relation is: \([\text{adj } A]^{-1} = A / \det(A)\)
For a 2x2 matrix \(A = \begin{vmatrix} a & b \\ c & d \end{vmatrix}\), the determinant is calculated as:
- \(\det(A) = ad - bc\)
In our case, let's calculate the determinant:
- \(\det(A) = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)\)
- \(\det(A) = \cos^2\theta + \sin^2\theta = 1\) (using \(\cos^2\theta + \sin^2\theta = 1\))
Since the determinant \(\det(A) = 1\), it implies for the inverse relationship:
- \([\text{adj } A]^{-1} = A\)
Thus, the correct answer is A.
Let's examine why other options are incorrect:
- -A: This is incorrect because \([\text{adj } A]^{-1} = A\) when \(\det(A) = 1\).
- -\(A^T\): This option does not relate to the properties of the inverse of the adjugate for unitary matrices.
- \(A^T\): This is incorrect because the inverse is the matrix itself due to \(\det(A) = 1\).
In summary, the matrix \(A\) is essentially a rotation matrix, which is orthogonal, and thus its determinant is 1. Therefore, \([\text{adj } A]^{-1} = A\), confirming the correct answer.
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