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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices1 Mark
MCQ
Choose the correct answer from the given four options. If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then $A - B$ is :
  • A
    $\text{I}$
  • B
    $0$
  • C
    $2\text{I}$
  • ✓
    $\frac{1}{2}\text{I}$
✓

Answer

Correct option: D.
$\frac{1}{2}\text{I}$
We have, $\text{B}=\begin{bmatrix}-\frac{1}{\pi}\cos^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&-\frac{1}{\pi}\tan^{-1}\pi\text{x}\end{bmatrix}$
and $\text{A}=\begin{bmatrix}\frac{1}{\pi}\sin^{-1}\text{x}\pi&\frac{1}{\pi}\tan^{-1}\frac{\text{x}}{\pi}\\\frac{1}{\pi}\sin^{-1}\frac{\text{x}}{\pi}&\frac{1}{\pi}\cot^{-1}\pi\text{x}\end{bmatrix}$
$\therefore\ \text{A}-\text{B}=\begin{bmatrix}\frac{1}{\pi}(\sin^{-1}\text{x}\pi+\cos^{-1}\text{x}\pi)&0\\0&\frac{1}{\pi}\big(\cot^{-1}\text{x}\pi+\tan^{-1}\pi\text{x}\big)\end{bmatrix}$
$=\begin{bmatrix}\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)&0\\0&\frac{1}{\pi}\Big(\frac{\pi}{2}\Big)\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\frac{1}{2}\text{I}$

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