If `F(x)=[("cos"x,-sin x,0),(sin x,cos x,0),(0,0,1)]` and `G(y)=[(cos y,0,sin y),(0,1,0),(-sin y,0,cos y)]`, then `[F(x) G(y)]^(-1)` is equal to

Options

A. `F(-x) G(-y)`
B. `G(-y) F(-x)`
C. `F(x^(-1))G(y^(-1))`
D. `G(y^(-1)) F(x^(-1))`

Explanation

To solve the problem, we need to find the inverse of the product of the matrices \( F(x) \) and \( G(y) \). The matrices are defined as follows:

\[ F(x) = \begin{pmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

\[ G(y) = \begin{pmatrix} \cos y & 0 & \sin y \\ 0 & 1 & 0 \\ -\sin y & 0 & \cos y \end{pmatrix} \]

We need to find \( [F(x) G(y)]^{-1} \).

### Step 1: Calculate \( G(y)^{-1} \)

To find the inverse of \( G(y) \), we first calculate its determinant.

\[ \text{det}(G(y)) = \cos y \cdot (1) \cdot \cos y + 0 + \sin y \cdot 0 \cdot (-\sin y) = \cos^2 y + \sin^2 y = 1 \]

Since the determinant is 1, we can find the adjoint of \( G(y) \) to get the inverse.

The adjoint of \( G(y) \) is obtained by taking the cofactor matrix and then transposing it.

Calculating the cofactor matrix:

- For \( G(y) \):
\[ \text{Cofactor}(G(y)) = \begin{pmatrix} \cos y & 0 & \sin y \\ 0 & 1 & 0 \\ -\sin y & 0 & \cos y \end{pmatrix} \]

The adjoint is:
\[ \text{adj}(G(y)) = \begin{pmatrix} \cos y & 0 & -\sin y \\ 0 & 1 & 0 \\ \sin y & 0 & \cos y \end{pmatrix} \]

Thus, the inverse of \( G(y) \) is:
\[ G(y)^{-1} = \text{adj}(G(y)) = \begin{pmatrix} \cos y & 0 & -\sin y \\ 0 & 1 & 0 \\ \sin y & 0 & \cos y \end{pmatrix} \]

### Step 2: Calculate \( F(x)^{-1} \)

Now, we will find the inverse of \( F(x) \).

The determinant of \( F(x) \) is:
\[ \text{det}(F(x)) = \cos^2 x + \sin^2 x = 1 \]

The adjoint of \( F(x) \) is:
\[ \text{Cofactor}(F(x)) = \begin{pmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Thus, the inverse of \( F(x) \) is:
\[ F(x)^{-1} = \text{adj}(F(x)) = \begin{pmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Calculate \( [F(x) G(y)]^{-1} \)

Using the property of inverses, we have:
\[ [F(x) G(y)]^{-1} = G(y)^{-1} F(x)^{-1} \]

Substituting the inverses we calculated:
\[ [F(x) G(y)]^{-1} = \begin{pmatrix} \cos y & 0 & -\sin y \\ 0 & 1 & 0 \\ \sin y & 0 & \cos y \end{pmatrix} \begin{pmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Perform the Matrix Multiplication

Calculating the product:
\[ [F(x) G(y)]^{-1} = \begin{pmatrix} \cos y \cos x + 0 + (-\sin y)(0) & \cos y \sin x + 0 + (-\sin y)(0) & 0 \\ 0 + 1 \cdot (-\sin x) + 0 & 0 + 1 \cdot \cos x + 0 & 0 \\ \sin y \cos x + 0 + \cos y(0) & \sin y \sin x + 0 + \cos y(0) & 0 \end{pmatrix} \]

This simplifies to:
\[ [F(x) G(y)]^{-1} = \begin{pmatrix} \cos y \cos x & \cos y \sin x & -\sin y \\ -\sin x & \cos x & 0 \\ \sin y \cos x & \sin y \sin x & \cos y \end{pmatrix} \]

### Final Result

Thus, we have:
\[ [F(x) G(y)]^{-1} = G(-y) F(-x) \]