For each real number x such that -1

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(C) Given $A(x) = \frac{1}{1-x^2} \begin{bmatrix} 1 & -x \ -x & 1 \end{bmatrix}$.We calculate the product $A(x)A(y)$:$A(x)A(y) = \frac{1}{1-x^2} \beg

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$A(z) = A(x)A(y)$

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(C) Given $A(x) = \frac{1}{1-x^2} \begin{bmatrix} 1 & -x \ -x & 1 \end{bmatrix}$.We calculate the product $A(x)A(y)$:$A(x)A(y) = \frac{1}{1-x^2} \begin{bmatrix} 1 & -x \ -x & 1 \end{bmatrix} \cdot \frac{1}{1-y^2} \begin{bmatrix} 1 & -y \ -y & 1 \end{bmatrix}$$= \frac{1}{(1-x^2)(1-y^2)} \begin{bmatrix} 1+xy & -(x+y) \ -(x+y) & 1+xy \end{bmatrix}$$= \frac{1+xy}{(1-x^2)(1-y^2)} \begin{bmatrix} 1 & -\frac{x+y}{1+xy} \ -\frac{x+y}{1+xy} & 1 \end{bmatrix}$Since $1-z^2 = 1 - (\frac{x+y}{1+xy})^2 = \frac{(1+xy)^2 - (x+y)^2}{(1+xy)^2} = \frac{1+2xy+x^2y^2 - x^2-2xy-y^2}{(1+xy)^2} = \frac{(1-x^2)(1-y^2)}{(1+xy)^2}$,we have $\frac{1}{1-z^2} = \frac{(1+xy)^2}{(1-x^2)(1-y^2)}$.Thus,$A(x)A(y) = \frac{1}{1-z^2} \begin{bmatrix} 1 & -z \ -z & 1 \end{bmatrix} = A(z)$.

For each real number $x$ such that $-1 < x < 1$,let $A(x)$ be the matrix $\frac{1}{1-x^2} \begin{bmatrix} 1 & -x \\ -x & 1 \end{bmatrix}$. If $z = \frac{x+y}{1+xy}$,then which of the following is true?

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