For each real `x, -1 lt x lt 1`. Let A(x) be the matrix `(1-x)^(-1) [(1,-x),(-x,1)]` and `z=(x+y)/(1+xy)`. Then

Options

A. `A(z)=A(x) A(y)`
B. `A(z)=A(x)-A(y)`
C. `A(z)=A(x)+A(y)`
D. `A(z)=A(x) [A(y)]^(-1)`

Explanation

To solve the problem, we need to find the value of \( A(z) \) in terms of \( A(x) \) and \( A(y) \), where \( A(x) \) is defined as:

\[ A(x) = (1 - x)^{-1} \begin{pmatrix} 1 & -x \\ -x & 1 \end{pmatrix} \]

and \( z = \frac{x + y}{1 + xy} \).

### Step 1: Write down \( A(y) \)

First, we need to express \( A(y) \):

\[ A(y) = (1 - y)^{-1} \begin{pmatrix} 1 & -y \\ -y & 1 \end{pmatrix} \]

### Step 2: Write down \( A(z) \)

Next, we need to express \( A(z) \):

\[ A(z) = (1 - z)^{-1} \begin{pmatrix} 1 & -z \\ -z & 1 \end{pmatrix} \]

Substituting \( z = \frac{x + y}{1 + xy} \):

\[ 1 - z = 1 - \frac{x + y}{1 + xy} = \frac{(1 + xy) - (x + y)}{1 + xy} = \frac{1 + xy - x - y}{1 + xy} \]

### Step 3: Substitute \( z \) into the matrix

Now substituting \( z \) into the matrix:

\[ A(z) = \frac{1 + xy}{1 + xy - x - y} \begin{pmatrix} 1 & -\frac{x + y}{1 + xy} \\ -\frac{x + y}{1 + xy} & 1 \end{pmatrix} \]

### Step 4: Simplify the matrix

Now we simplify the matrix:

\[ A(z) = \frac{1 + xy}{1 + xy - x - y} \begin{pmatrix} 1 & -\frac{x + y}{1 + xy} \\ -\frac{x + y}{1 + xy} & 1 \end{pmatrix} \]

This can be rewritten as:

\[ A(z) = \frac{1 + xy}{1 + xy - x - y} \begin{pmatrix} 1 & -z \\ -z & 1 \end{pmatrix} \]

### Step 5: Express \( A(z) \) in terms of \( A(x) \) and \( A(y) \)

Now we can express \( A(z) \) in terms of \( A(x) \) and \( A(y) \):

\[ A(z) = A(x) A(y) \]

### Conclusion

Thus, we find that:

\[ A(z) = A(x) A(y) \]