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Definition of a Determinant

If A=1π[ si...

Question

If $A = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦, B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$ then $A - B$ is equal to

A

I

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B

0

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C

2 I

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D

\frac{1}{2} I

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Solution

The correct option is C $\frac{1}{2} I$
We have, $A - B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ - \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow π (A - B) = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow π (A - B) = [\begin{matrix} {sin}^{- 1} (π x) + {cos}^{- 1} (π x) & 0 0 & {tan}^{- 1} (π x) + {cot}^{- 1} (π x) \end{matrix}]$ , Subtract the corresponding elements
$\Rightarrow π (A - B) = ⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{π}{2} & 0 0 & \frac{π}{2} \end{matrix} ⎤ ⎥ ⎥ ⎦$ , use the identities ${sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2} = {tan}^{- 1} x + {cot}^{- 1} x$
$\Rightarrow π (A - B) = \frac{π}{2} [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] = \frac{π}{2} I$ , take $\frac{π}{2}$ common
$∴ A - B = \frac{1}{2} I$

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