Question

For each real number x such that – 1 < x < 1, let A (x) be the matrix ${{\left( 1-x \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]$ and $z=\dfrac{x+y}{1+xy},$ then,
$\left( a \right)A\left( z \right)=A\left( x \right)+A\left( y \right)$
$\left( b \right)A\left( z \right)=A\left( x \right)+{{\left[ A\left( y \right) \right]}^{-1}}$
$\left( c \right)A\left( z \right)=A\left( x \right)A\left( y \right)\div \sqrt{1+xy}$
$\left( d \right)A\left( z \right)=A\left( x \right)-A\left( y \right)$

Answer 1

To solve this question, we will first of all note the value of A(x) and then replace x by z to obtain A(z) other than A(x). Finally, we will open $z=\dfrac{x+y}{1+xy}$ inside A(z) and then try to obtain $A\left( x \right)={{\left( 1-x \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]$ and $A\left( y \right)={{\left( 1-y \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]$ inside to make the proper substitution.

Complete step by step answer:
We are given that $A\left( x \right)={{\left( 1-x \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]$ and $z=\dfrac{x+y}{1+xy}.$ Also, - 1 < x < 1. Replacing x by z in A(x), we get,

1 & -z \\\ -z & 1 \\\ \end{matrix} \right]$$ Putting the value of $$z=\dfrac{x+y}{1+xy}$$ in the above equation, we get, $$\Rightarrow A\left( z \right)={{\left( 1-\left( \dfrac{x+y}{1+xy} \right) \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & \dfrac{-\left( x+y \right)}{\left( 1+xy \right)} \\\ \dfrac{-\left( x+y \right)}{\left( 1+xy \right)} & 1 \\\ \end{matrix} \right]$$ Taking $$\dfrac{1}{1+xy}$$ common from the matrix in the RHS of the above equation, we have, $$\Rightarrow A\left( z \right)={{\left( 1-\dfrac{x+y}{1+xy} \right)}^{\dfrac{-1}{2}}}\dfrac{1}{1+xy}\left[ \begin{matrix} 1+xy & -x-y \\\ -x-y & 1+xy \\\ \end{matrix} \right]$$ Now splitting the matrix $$\left[ \begin{matrix} 1+xy & -x-y \\\ -x-y & 1+xy \\\ \end{matrix} \right]$$ as $$\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right],$$ we get, $$\Rightarrow A\left( z \right)={{\left( 1-\dfrac{x+y}{1+xy} \right)}^{\dfrac{-1}{2}}}\dfrac{1}{1+xy}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]$$ Now solving the term $${{\left( 1-\dfrac{x+y}{1+xy} \right)}^{\dfrac{-1}{2}}}$$ we get, $$\Rightarrow A\left( z \right)={{\left( \dfrac{1+xy-x-y}{1+xy} \right)}^{\dfrac{-1}{2}}}\dfrac{1}{\left( 1+xy \right)}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]$$ $$\Rightarrow A\left( z \right)={{\left( \dfrac{\left( 1-x \right)\left( 1-y \right)}{1+xy} \right)}^{\dfrac{-1}{2}}}\dfrac{1}{\left( 1+xy \right)}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]$$ Now because $$A\left( x \right)={{\left( 1-x \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]$$ and $$A\left( y \right)={{\left( 1-y \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right],$$then substituting this in the above, we get, $$\Rightarrow A\left( z \right)=\dfrac{{{\left( 1-x \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]{{\left( 1-y \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]}{{{\left( 1+xy \right)}^{\dfrac{-1}{2}}}{{\left( 1+xy \right)}^{1}}}$$ $$\Rightarrow A\left( z \right)=\dfrac{{{\left( 1-x \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]{{\left( 1-y \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]}{{{\left( 1+xy \right)}^{\dfrac{1}{2}}}}$$ Using the value of A(x) and A(y), we get, $$\Rightarrow A\left( z \right)=\dfrac{1}{{{\left( 1+xy \right)}^{\dfrac{1}{2}}}}A\left( x \right)A\left( y \right)$$ $$\Rightarrow A\left( z \right)=A\left( x \right)A\left( y \right)\div \sqrt{1+xy}$$ **So, the correct answer is “Option C”.** **Note:** The possibility of the confusion in this question can be at the part where we have shifted $$\dfrac{\left[ \left( 1-x \right)\left( 1-y \right) \right]}{{{\left( 1+xy \right)}^{\dfrac{-1}{2}}}}$$ as $$\left( \dfrac{1}{1+xy} \right)\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]$$ to $$\dfrac{{{\left( 1-x \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]{{\left( 1-y \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]}{{{\left( 1+xy \right)}^{\dfrac{-1}{2}}}{{\left( 1+xy \right)}^{1}}}.$$ It is not wring, it is correct as even if the matrix multiplication is not commutative then also separately $$\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]$$ and $$\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]$$ by a term $${{\left( 1-y \right)}^{\dfrac{-1}{2}}}$$ works as $${{\left( 1-y \right)}^{\dfrac{-1}{2}}}$$ is a scalar and not a matrix. Clearly, we have not changed the position of both the matrices using $$\left[ \begin{matrix} 1 & -y \\\ -y & 1 \\\ \end{matrix} \right]$$ and $$\left[ \begin{matrix} 1 & -x \\\ -x & 1 \\\ \end{matrix} \right]$$ with the change in order would be wrong.