Q. For each real$\ x,\ - 1\ < x < 1$. Let $A(x)$be the matrix $(1 - x)^{- 1}\begin{bmatrix} 1 & - x \\ - x & 1 \end{bmatrix}$ and $\mathcal{z =}\frac{x + y}{1 + xy}$. Then
A) $A\left( \mathcal{z} \right) = A(x)A(y)$
B) $A\left( \mathcal{z} \right) = A(x) - A(y)$
C) $A\left( \mathcal{z} \right) = A(x) + A(y)$
D) $A\left( \mathcal{z} \right) = A(x)\left\lbrack A(y) \right\rbrack^{- 1}$
Explanation
(a)
$A(x)A(y) = (1 - x)^{- 1}(1 - y)^{- 1}\begin{bmatrix}
1 & - x \\
- x & 1
\end{bmatrix}\begin{bmatrix}
1 & - y \\
- y & 1
\end{bmatrix}$
$= \left( 1 + xy - (x + y) \right)^{- 1}\begin{bmatrix}
1 + xy & - (x + y) \\
- (x + y) & 1 + xy
\end{bmatrix}$
$= \left( 1 - \frac{(x + y)}{1 + xy} \right)^{- 1}\begin{bmatrix}
1 & - \frac{x + y}{1 + xy} \\
- \frac{x + y}{1 + xy} & 1
\end{bmatrix} = A(\mathcal{z)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.