Question

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

• A. $A\left(z\right)=A\left(x\right)+A\left(y\right)$
• B. $A\left(z\right)=A\left(x\right)+\left[A\right(y){]}^{-1}$
• C. $A\left(z\right)=A\left(x\right)A\left(y\right)$
• D. $A\left(z\right)=A\left(x\right)-A\left(y\right)$

Answer 1

The correct option is C $A\left(z\right)=A\left(x\right)A\left(y\right)$

$A\left(z\right)=A\left(\frac{x+y}{1+xy}\right)=\left[\frac{1+xy}{(1-x)(1-y)}\right]\left[\begin{array}{cc}1& -\left(\frac{x+y}{1+xy}\right)\\ -\left(\frac{x+y}{1+xy}\right)& 1\end{array}\right]\therefore A\left(x\right).A\left(y\right)=A\left(z\right).$