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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices1 Mark
MCQ
If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&\cot^{-1}(\pi\text{x})\end{bmatrix}, \text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&-\tan^{-1}(\pi\text{x})\end{bmatrix},$ then $A - B$ is equal to :
  • ✓
    $I$
  • B
    $0$
  • C
    $2I$
  • D
    $\frac{1}{2}\text{I}$
✓

Answer

Correct option: A.
$I$
Given $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&\cot^{-1}(\pi\text{x})\end{bmatrix}$
$\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&-\tan^{-1}(\pi\text{x})\end{bmatrix}$
$\text{A}-\text{B}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)+\cos^{-1}(\pi\text{x})&0\\0&\cot^{-1}(\pi\text{x})+\tan^{-1}(\pi\text{x})\end{bmatrix}$
$=\frac{1}{\pi}\begin{bmatrix}\frac{\pi}{2}&0\\0&\frac{\pi}{2}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac{1}{2}\text{I}$

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