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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
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For each real numbers $x$ such that $-1 < x < 1$ let $A(x)$ be the matrix.$(1-x)^{-\large\frac{1}{2}}\begin{bmatrix}1&-x\\-x&1\end{bmatrix}$ and $z=\large\frac{x+y}{1+xy}$.Then

$\begin{array}{1 1}(A)\;A(z)=A(x)+A(y)&(B)\;A(z)=A(x)[A(y)]^{-1}\\(C)\;A(z)=A(x)A(y)+\sqrt{1+xy}&(D)\;A(z)=A(x)-A(y)\end{array}$

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1 Answer

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$A(z)=A\big(\large\frac{x+y}{1+xy}\big)$
$\Rightarrow \big(1-\large\frac{x+y}{1+xy}\big)^{-\large\frac{1}{2}}\begin{bmatrix}1&-\large\frac{x+y}{1+xy}\\\large\frac{-(x+y)}{1+xy}&1\end{bmatrix}$
$A(x)A(y)=(1-x)^{-\large\frac{1}{2}}\begin{bmatrix}1&-x\\-x&1\end{bmatrix}(1-y)^{-\large\frac{1}{2}}\begin{bmatrix}1&-y\\-y&1\end{bmatrix}$
$\Rightarrow (1+xy-x-y)^{\large\frac{-1}{2}}\begin{bmatrix}1+xy&-(x+y)\\-(x+y)&1+xy\end{bmatrix}$
$A(x)A(y)=(1+xy-x-y)^{\large\frac{-1}{2}}(1+xy)$$\times \begin{bmatrix}1&-\large\frac{(x+y)}{1+xy}\\-\large\frac{(x+y)}{1+xy}&1\end{bmatrix}$
$\large\frac{A(x)A(y)}{\sqrt{1+xy}}=\big(\large\frac{1+xy-x-y}{1+xy}\big)^{-\large\frac{1}{2}}$$\begin{bmatrix}1&-\large\frac{(x+y)}{1+xy}\\-\large\frac{(x+y)}{1+xy}&1\end{bmatrix}$
Hence (C) is the correct answer.
answered Apr 23, 2014 by sreemathi.v
 

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