question_answer

If \[A=\frac{1}{\pi }\left[ \begin{matrix}    {{\sin }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right)  \\    {{\sin }^{-1}}\left( \frac{x}{\pi } \right) & {{\cot }^{-1}}(\pi x)  \\ \end{matrix} \right]\] and \[A=\frac{1}{\pi }\left[ \begin{matrix}    -{{\cos }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right)  \\    {{\sin }^{-1}}\left( \frac{x}{\pi } \right) & -ta{{n}^{-1}}(x\pi )  \\ \end{matrix} \right],\] Then, find the value of \[A\,-B\] in terms of identity matrix I.

Answer:

Consider \[A\,-B=\frac{1}{\pi }\left\{ \left[ \begin{matrix}    {{\sin }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right)  \\    {{\sin }^{-1}}\left( \frac{x}{\pi } \right) & {{\cot }^{-1}}(x\pi )  \\ \end{matrix} \right] \right.\]                         \[\left. -\left[ \begin{matrix}    -{{\cos }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right)  \\    {{\sin }^{-1}}\left( \frac{x}{\pi } \right) & -{{\tan }^{-1}}(\pi x)  \\ \end{matrix} \right] \right\}\]  \[=\frac{1}{\pi }\left[ \begin{matrix}    {{\sin }^{-1}}(x\pi )+{{\cos }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right)-{{\tan }^{-1}}\left( \frac{x}{\pi } \right)  \\    {{\sin }^{-1}}\left( \frac{x}{\pi } \right)-{{\sin }^{-1}}\left( \frac{x}{\pi } \right) & {{\cot }^{-1}}(x\pi )+{{\tan }^{-1}}(\pi x)  \\ \end{matrix} \right]\] \[=\frac{1}{\pi }\left[ \begin{matrix}    \frac{\pi }{2} & 0  \\    0 & \frac{\pi }{2}  \\ \end{matrix} \right]\] \[\left[ \begin{align}   & \because {{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2};x\in []-1,1 \\  & \text{and}\,\,{{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2},x\in R \\ \end{align} \right]\]   \[=\left[ \begin{matrix}    \frac{1}{2} & 0  \\    0 & \frac{1}{2}  \\ \end{matrix} \right]=\frac{1}{2}l\]