Test Index

UPSEE Solved Model Paper 4

© examsnet.com

Question : 144 of 150

Marks: +1, -0

If A =

\frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\pi/x) \\ \sin^{-1}(\pi/x) & \tan^{-1}(\pi x) \end{bmatrix}

\frac{1}{\pi} \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\pi/x) \\ \sin^{-1}(\pi/x) & -\tan^{-1}(\pi x) \end{bmatrix}

then

I
0
2I
1/2 I

Solution:

A - B =

\frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\pi/x) \\ \sin^{-1}(\pi/x) & \tan^{-1}(\pi x) \end{bmatrix}

-

\frac{1}{\pi} \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\pi/x) \\ \sin^{-1}(\pi/x) & -\tan^{-1}(\pi x) \end{bmatrix}

⇒ π (A - B) =

\frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\pi/x) \\ \sin^{-1}(\pi/x) & \tan^{-1}(\pi x) \end{bmatrix}

-

\begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\pi/x) \\ \sin^{-1}(\pi/x) & -\tan^{-1}(\pi x) \end{bmatrix}

⇒ π (A - B) =

\begin{bmatrix} \sin^{-1}(\pi x) + \cos^{-1}(\pi x) & 0 \\ 0 & \tan^{-1}(\pi x) + \cot^{-1}(\pi x) \end{bmatrix}

Subtract the corresponding elements

⇒ π (A - B) =

\begin{bmatrix} \frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2} \end{bmatrix}

, use the identities

\sin^{-1} x + \cos^{-1} x

=

\frac{\pi}{2}

=

\tan^{-1} x + \cot^{-1} x

⇒ π (A - B) =

\frac{\pi}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

=

\frac{\pi}{2} I

, take

\frac{\pi}{2}

common

∴ A - B =

\frac{1}{2} I

© examsnet.com

Go to Question:

Prev Question

Next Question