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

Hint: 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,
\[A\left( z \right)={{\left( 1-z \right)}^{\dfrac{-1}{2}}}\left[ \begin{matrix}
   1 & -z \\
   -z & 1 \\
\end{matrix} \right]\]
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.