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 find \([F(x) G(y)]^{-1}\), we can use the property of the inverse of the product of matrices. The property states that \((AB)^{-1} = B^{-1} A^{-1}\).

Let's go through the steps to find \([F(x) G(y)]^{-1}\):

### Step 1: Write down the matrices \(F(x)\) and \(G(y)\)
Given:
\[ F(x) = \begin{pmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
\[ G(y) = \begin{pmatrix} \cos y & 0 & \sin y \\ 0 & 1 & 0 \\ -\sin y & 0 & \cos y \end{pmatrix} \]

### Step 2: Find the inverses \(F(x)^{-1}\) and \(G(y)^{-1}\)
The inverse of a rotation matrix can be found by taking the transpose of the matrix. Thus:
\[ F(x)^{-1} = F(-x) = \begin{pmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Similarly, for \(G(y)\):
\[ G(y)^{-1} = G(-y) = \begin{pmatrix} \cos(-y) & 0 & \sin(-y) \\ 0 & 1 & 0 \\ -\sin(-y) & 0 & \cos(-y) \end{pmatrix} = \begin{pmatrix} \cos y & 0 & -\sin y \\ 0 & 1 & 0 \\ \sin y & 0 & \cos y \end{pmatrix} \]

### Step 3: Use the property of inverses
Now, using the property of inverses:
\[ [F(x) G(y)]^{-1} = G(y)^{-1} F(x)^{-1} \]

### Step 4: Substitute the inverses
Substituting the inverses we found:
\[ [F(x) G(y)]^{-1} = G(-y) F(-x) \]
\[ = \begin{pmatrix} \cos y & 0 & -\sin y \\ 0 & 1 & 0 \\ \sin y & 0 & \cos y \end{pmatrix} \begin{pmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Multiply the matrices
Now, we perform the matrix multiplication:
\[ = \begin{pmatrix} \cos y \cos x + 0 \cdot (-\sin x) + (-\sin y) \cdot 0 & \cos y \sin x + 0 \cdot \cos x + (-\sin y) \cdot 0 & \cos y \cdot 0 + 0 \cdot 0 + (-\sin y) \cdot 1 \\ 0 \cdot \cos x + 1 \cdot (-\sin x) + 0 \cdot 0 & 0 \cdot \sin x + 1 \cdot \cos x + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 1 \\ \sin y \cdot \cos x + 0 \cdot (-\sin x) + \cos y \cdot 0 & \sin y \cdot \sin x + 0 \cdot \cos x + \cos y \cdot 0 & \sin y \cdot 0 + 0 \cdot 0 + \cos y \cdot 1 \end{pmatrix} \]
This simplifies to:
\[ = \begin{pmatrix} \cos y \cos x & \cos y \sin x & -\sin y \\ -\sin x & \cos x & 0 \\ \sin y \cos x & \sin y \sin x & \cos y \end{pmatrix} \]

### Final Answer
Thus, the final result is:
\[ [F(x) G(y)]^{-1} = \begin{pmatrix} \cos y \cos x & \cos y \sin x & -\sin y \\ -\sin x & \cos x & 0 \\ \sin y \cos x & \sin y \sin x & \cos y \end{pmatrix} \]