For each real x, −1<x<1. Let A(x) be the matrix (1−x)−11−x−x1 and z=x+y1+xy. Then

A(x)A(y)=(1−x)−1(1−y)−11−x−x11−y−y1 =(1+xy−(x+y))−11+xy−(x+y)−(x+y)1+xy =1−x+y1+xy−11−x+y1+xy−x+y1+xy1=A(z)

For each real x, −1<x<1. Let A(x) be the matrix (1−x)−11−x−x1 and z=x+y1+xy. Then

A(x)A(y)=(1−x)−1(1−y)−11−x−x11−y−y1 =(1+xy−(x+y))−11+xy−(x+y)−(x+y)1+xy =1−x+y1+xy−11−x+y1+xy−x+y1+xy1=A(z)

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For each real $x, - 1 < x < 1$ . Let A(x) be the matrix $(1 - x)^{- 1} [\begin{array}{cc} 1 & - x \\ - x & 1 \end{array}] and z = \frac{x + y}{1 + xy}$ . Then

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

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

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

$A (z) = A (x) [A (y)]^{- 1}$

answer is A.

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Detailed Solution

$\begin{array}{c} A (x) A (y) = (1 - x)^{- 1} (1 - y)^{- 1} [\begin{array}{cc} 1 & - x \\ - x & 1 \end{array}] [\begin{array}{cc} 1 & - y \\ - y & 1 \end{array}] \\ = (1 + xy - (x + y))^{- 1} [\begin{array}{cc} 1 + xy & - (x + y) \\ - (x + y) & 1 + xy \end{array}] \\ = {(1 - \frac{x + y}{1 + xy})}^{- 1} [\begin{array}{cc} 1 & - \frac{x + y}{1 + xy} \\ - \frac{x + y}{1 + xy} & 1 \end{array}] = A (z) \end{array}$