Let's find the inverse of the given matrix step-by-step. The given matrix is:

To find the inverse of a matrix \(A\), denoted as \(A^{-1}\), we use the formula:

where \(\det(A)\) is the determinant of matrix \(A\), and \(\text{adj}(A)\) is the adjoint of matrix \(A\).

Calculating the Determinant of the Matrix

First, we need to calculate the determinant of matrix \(A\). We can expand along the third row because it has two zeros, which simplifies the calculation.

where \(C_{ij}\) is the cofactor of the element in the i-th row and j-th column.

We only need to calculate \(C_{33}\):

where \(M_{33}\) is the minor obtained by deleting the 3rd row and 3rd column of \(A\).

So, \(C_{33} = (-1)^6 \cdot 1 = 1 \cdot 1 = 1\).

Therefore, the determinant of \(A\) is:

Since \(\det(A) = 1 \neq 0\), the inverse of the matrix \(A\) exists.

Calculating the Adjoint of the Matrix

The adjoint of \(A\), \(\text{adj}(A)\), is the transpose of the matrix of cofactors of \(A\).

Let \(C\) be the matrix of cofactors, where \(C_{ij}\) is the cofactor of the element \(a_{ij}\) in matrix \(A\).

Let's calculate each cofactor:

• \(C_{11} = (-1)^{1+1} \det \left( {\begin{array}{*{20}{c}} {\cos \theta }&0\\ 0&1 \end{array}} \right) = (1)(\cos \theta \cdot 1 - 0 \cdot 0) = \cos \theta\)
• \(C_{12} = (-1)^{1+2} \det \left( {\begin{array}{*{20}{c}} { - \sin \theta }&0\\ 0&1 \end{array}} \right) = (-1)((-\sin \theta) \cdot 1 - 0 \cdot 0) = \sin \theta\)
• \(C_{13} = (-1)^{1+3} \det \left( {\begin{array}{*{20}{c}} { - \sin \theta }&{\cos \theta }\\ 0&0 \end{array}} \right) = (1)((-\sin \theta) \cdot 0 - \cos \theta \cdot 0) = 0\)
• \(C_{21} = (-1)^{2+1} \det \left( {\begin{array}{*{20}{c}} {\sin \theta }&0\\ 0&1 \end{array}} \right) = (-1)(\sin \theta \cdot 1 - 0 \cdot 0) = -\sin \theta\)
• \(C_{22} = (-1)^{2+2} \det \left( {\begin{array}{*{20}{c}} {\cos \theta }&0\\ 0&1 \end{array}} \right) = (1)(\cos \theta \cdot 1 - 0 \cdot 0) = \cos \theta\)
• \(C_{23} = (-1)^{2+3} \det \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }\\ 0&0 \end{array}} \right) = (-1)(\cos \theta \cdot 0 - \sin \theta \cdot 0) = 0\)
• \(C_{31} = (-1)^{3+1} \det \left( {\begin{array}{*{20}{c}} {\sin \theta }&0\\ {\cos \theta }&0 \end{array}} \right) = (1)(\sin \theta \cdot 0 - 0 \cdot \cos \theta) = 0\)
• \(C_{32} = (-1)^{3+2} \det \left( {\begin{array}{*{20}{c}} {\cos \theta }&0\\ { - \sin \theta }&0 \end{array}} \right) = (-1)(\cos \theta \cdot 0 - 0 \cdot (-\sin \theta)) = 0\)
• \(C_{33} = (-1)^{3+3} \det \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }\\ { - \sin \theta }&{\cos \theta } \end{array}} \right) = (1)(\cos \theta \cdot \cos \theta - \sin \theta \cdot (-\sin \theta)) = \cos^2 \theta + \sin^2 \theta = 1\)

The matrix of cofactors \(C\) is:

The adjoint of \(A\) is the transpose of the cofactor matrix \(C\):

Finding the Inverse Matrix

Now, we use the formula \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\).

Since \(\det(A) = 1\), the inverse is simply the adjoint matrix:

Comparing with Options

Let's compare our calculated inverse with the given options:

The calculated inverse matches Option 1.

Revision Table: Matrix Inverse Calculation

Additional Information: Rotation Matrix and Inverse

The given matrix \(A\) is a special type of matrix. It represents a 2D rotation by an angle \(\theta\) in the xy-plane, combined with no change in the z-direction. Specifically, the upper-left 2x2 part is a standard 2D rotation matrix:

The inverse of a rotation matrix \(R(\theta)\) is its transpose, \(R(\theta)^T\). The transpose of \(R(\theta)\) is:

which corresponds to a rotation by \(-\theta\).

For the given 3x3 matrix \(A\), the inverse operation would be rotating back by \(-\theta\) in the xy-plane, while still keeping the z-component unchanged. This suggests the inverse matrix should have the 2D rotation by \(-\theta\) in the upper-left block and a 1 in the (3,3) position.

The matrix for a 2D rotation by \(-\theta\) is obtained by replacing \(\theta\) with \(-\theta\) in the rotation matrix formula:

Combining this with the unchanged z-component, the inverse matrix is expected to be:

This matches the result obtained through the determinant and adjoint method, confirming that the matrix represents a transformation involving rotation.