Question

For each real x such that –1 < x < 1 . Let A (x) denote the matrix (1 – x)^–1/2$\begin{pmatrix} 1 & - x \

• x & 1 \end{pmatrix}$and A(x) A(y) = k A(z) where x, y Ī R, –1 \< x, y \< 1 and z =$\frac{x + y}{1 + xy}$ then k is –

• A. $\frac{1}{\sqrt{1 + xy}}$
• B. $\frac{x}{\sqrt{1 + xy}}$
• C. $\sqrt{1 + xy}$
• D. $\frac{\sqrt{1 + xy}}{x}$

Answer 1

Answer: (C)

$\sqrt{1 + xy}$

Answer 2

A(x) A(y) = (1 – y)^–1/2 $\begin{pmatrix} 1 & - x \

• x & 1 \end{pmatrix}$$\begin{pmatrix} 1 & - y \
• y & 1 \end{pmatrix}$

= (1 – x – y + xy)^–1/2 $\begin{bmatrix} 1 + xy & - (x + y) \

• (x + y) & 1 + xy \end{bmatrix}$

= $\frac { 1 + x y } { \sqrt { 1 - ( x + y ) + ( x y ) } }$ $\begin{bmatrix} 1 & \frac{- (x + y)}{1 + xy} \\ \frac{- (x + y)}{1 + xy} & 1 \end{bmatrix}$

=$\begin{bmatrix} 1 & \frac{- (x + y)}{1 + xy} \\ \frac{- (x + y)}{1 + xy} & 1 \end{bmatrix}$

= $\sqrt{1 + xy}$ A(z)

Ž k = $\sqrt{1 + xy}$.