For each real number x such that − 1 < x < 1 , l e t A ( x ) b e t h e m a t r i x ( 1 − x ) − 1 [ 1 − x − x 1 ] a n d z = x + y 1 + x y T h e n , Options: A . &nbspA(z)=A(x)+A(y) B . &nbspA(z)=A(x)+[A(y)]−1 C . &nbspA(z)=A(x)A(y) D . &nbspA(z)=A(x)–A(y)

Answer: Option C : C A ( z ) = A ( x + y 1 + x y ) = [ 1 + x y ( 1 − x ) ( 1 − y ) ] ⎡ ⎢ ⎣ 1 − ( x + y 1 + x y ) − ( x + y 1 + x y ) 1 ⎤ ⎥ ⎦ ∴ A ( x ) . A ( y ) = A ( z )

For each real number x such that −1<x<1,let A(x) be

Question

For each real number x such that

- 1 < x < 1, l e t A (x) b e t h e m a t r i x (1 - x)^{- 1} [\begin{matrix} 1 & - x - x & 1 \end{matrix}] a n d z = \frac{x + y}{1 + x y} T h e n,

Options:

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

B . A(z)=A(x)+[A(y)]−1

C . A(z)=A(x)A(y)

D . A(z)=A(x)–A(y)

Answer: Option C
:
C

A (z) = A (\frac{x + y}{1 + x y}) = [\frac{1 + x y}{(1 - x) (1 - y)}] ⎡ ⎢ ⎣ \begin{matrix} 1 & - (\frac{x + y}{1 + x y}) - (\frac{x + y}{1 + x y}) & 1 \end{matrix} ⎤ ⎥ ⎦

∴ A (x) . A (y) = A (z)

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