If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to

Options

A. `A(-alpha, -beta)`
B. `A(-alpha, beta)`
C. `A(alpha, -beta)`
D. `A(alpha, beta)`

Explanation

To find the inverse of the matrix \( A(\alpha, \beta) \), we will follow these steps:

### Step 1: Write down the matrix \( A(\alpha, \beta) \)
The given matrix is:
\[ A(\alpha, \beta) = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^{\beta} \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A(\alpha, \beta) \)
The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \text{det}(A) = \cos \alpha \cdot (\cos \alpha \cdot e^{\beta} - 0) - \sin \alpha \cdot (-\sin \alpha \cdot e^{\beta} - 0) + 0 \]
This simplifies to:
\[ \text{det}(A) = \cos^2 \alpha \cdot e^{\beta} + \sin^2 \alpha \cdot e^{\beta} = e^{\beta} (\cos^2 \alpha + \sin^2 \alpha) = e^{\beta} \]

### Step 3: Find the adjoint of \( A(\alpha, \beta) \)
The adjoint of a matrix is the transpose of its cofactor matrix. We will calculate the cofactor matrix first.

The cofactor matrix is calculated as follows:
1. For the element at (1,1): \( C_{11} = e^{\beta} \)
2. For the element at (1,2): \( C_{12} = -0 = 0 \)
3. For the element at (1,3): \( C_{13} = \sin \alpha \cdot e^{\beta} \)
4. For the element at (2,1): \( C_{21} = -\sin \alpha \cdot e^{\beta} \)
5. For the element at (2,2): \( C_{22} = e^{\beta} \)
6. For the element at (2,3): \( C_{23} = 0 \)
7. For the element at (3,1): \( C_{31} = 0 \)
8. For the element at (3,2): \( C_{32} = 0 \)
9. For the element at (3,3): \( C_{33} = \cos^2 \alpha + \sin^2 \alpha = 1 \)

Thus, the cofactor matrix is:
\[ \text{Cofactor}(A) = \begin{pmatrix} e^{\beta} & 0 & \sin \alpha \cdot e^{\beta} \\ -\sin \alpha \cdot e^{\beta} & e^{\beta} & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Now, take the transpose to get the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} e^{\beta} & -\sin \alpha \cdot e^{\beta} & 0 \\ 0 & e^{\beta} & 0 \\ \sin \alpha \cdot e^{\beta} & 0 & 1 \end{pmatrix} \]

### Step 4: Calculate the inverse using the formula
The formula for the inverse of a matrix is:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]
Substituting the values we found:
\[ A^{-1} = \frac{1}{e^{\beta}} \begin{pmatrix} e^{\beta} & -\sin \alpha \cdot e^{\beta} & 0 \\ 0 & e^{\beta} & 0 \\ \sin \alpha \cdot e^{\beta} & 0 & 1 \end{pmatrix} \]
This simplifies to:
\[ A^{-1} = \begin{pmatrix} 1 & -\sin \alpha & 0 \\ 0 & 1 & 0 \\ \sin \alpha & 0 & e^{-\beta} \end{pmatrix} \]

### Step 5: Final Result
Thus, the inverse of the matrix \( A(\alpha, \beta) \) is:
\[ A^{-1}(\alpha, \beta) = \begin{pmatrix} \cos(-\alpha) & \sin(-\alpha) & 0 \\ -\sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & e^{-\beta} \end{pmatrix} \]