Question

For each real number $x$ such that $-1<x<1$, let $A\left(x\right)$ be the matrix $(1-x{)}^{-1/2}\left[\begin{array}{cc}1& -x\\ -x& 1\end{array}\right]$ 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. $A\left(z\right)=A\left(x\right)A\left(y\right)÷\sqrt{1+xy}$
• D. $A\left(z\right)=A\left(x\right)-A\left(y\right)$

Answer 1

The correct option is C $A\left(z\right)=A\left(x\right)A\left(y\right)÷\sqrt{1+xy}$

Given $A\left(x\right)={(1-x)}^{-\frac{1}{2}}\left[\begin{array}{cc}1& -x\\ -x& 1\end{array}\right]$

and $z=\frac{x+y}{1+xy}$

So we have, $A\left(z\right)={(1-z)}^{-\frac{1}{2}}\left[\begin{array}{cc}1& -z\\ -z& 1\end{array}\right]$

Putting value of z we get, $\Rightarrow A\left(z\right)={(1-\frac{x+y}{1+xy})}^{-\frac{1}{2}}\left[\begin{array}{cc}1& -\frac{x+y}{1+xy}\\ -\frac{x+y}{1+xy}& 1\end{array}\right]$

$\Rightarrow A\left(z\right)={(1-\frac{x+y}{1+xy})}^{-\frac{1}{2}}\frac{1}{1+xy}\left[\begin{array}{cc}1+xy& -x-y\\ -x-y& 1+xy\end{array}\right]$

$\Rightarrow A\left(z\right)={(1-\frac{x+y}{1+xy})}^{-\frac{1}{2}}\frac{1}{1+xy}\left[\begin{array}{cc}1& -x\\ -x& 1\end{array}\right]\left[\begin{array}{cc}1& -y\\ -y& 1\end{array}\right]$

$\Rightarrow A\left(z\right)={\left(\frac{1-x-y+xy}{1+xy}\right)}^{-\frac{1}{2}}\frac{1}{1+xy}\left[\begin{array}{cc}1& -x\\ -x& 1\end{array}\right]\left[\begin{array}{cc}1& -y\\ -y& 1\end{array}\right]$

$\Rightarrow A\left(z\right)={\left(\frac{(1-x)(1-y)}{1+xy}\right)}^{-\frac{1}{2}}\frac{1}{1+xy}\left[\begin{array}{cc}1& -x\\ -x& 1\end{array}\right]\left[\begin{array}{cc}1& -y\\ -y& 1\end{array}\right]$

$\Rightarrow A\left(z\right)=\frac{1}{{(1+xy)}^{\frac{1}{2}}}A\left(x\right)A\left(y\right)$