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Definition of a Determinant

If A = 1π[ ...

Question

If $A = \frac{1}{π} ⎡ ⎢ ⎢ ⎣ \begin{matrix} s i n^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦, B = ⎡ ⎢ ⎢ ⎣ \begin{matrix} - c o s^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & t a n^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦$ then A - B is equal to

A

0

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B

2I

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C

I

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D

\frac{1}{2} I

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Solution

The correct option is D $\frac{1}{2} I$
$A = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{x}{π}) {sin}^{- 1} (\frac{x}{π}) & {cot}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{x}{π}) {sin}^{- 1} (\frac{x}{π}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$A - B = \frac{1}{π} [\begin{matrix} {sin}^{- 1} (π x) + {cos}^{- 1} (π x) & 0 0 & {cot}^{- 1} (π x) + {tan}^{- 1} (π x) \end{matrix}]$
$A - B = \frac{1}{π} [\begin{matrix} π / 2 & 0 0 & π / 2 \end{matrix}]$
$A - B = \frac{1}{π} * \frac{π}{2} [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$
$A - B = \frac{1}{2} I$

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Similar questions

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