
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
561.3k+ views
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.
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.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Which is the Longest Railway Platform in the world?

India Manned Space Mission Launch Target Month and Year 2025 Update

Which of the following pairs is correct?

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Plot a graph between potential difference V and current class 12 physics CBSE

