If A = frac{1}{pi} begin{bmatrix} sin^{-1}(xp | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given matrices are $A = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(x\pi) & \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) \ \frac{1}{\pi} \sin^{-1}(\frac{x}

Vedclass · Accepted Answer

$\frac{1}{2}I$

Vedclass · Answer

(D) Given matrices are $A = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(x\pi) & \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) \ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & \frac{1}{\pi} \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \begin{bmatrix} -\frac{1}{\pi} \cos^{-1}(x\pi) & \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) \ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & -\frac{1}{\pi} 	an^{-1}(\pi x) \end{bmatrix}$.Subtracting $B$ from $A$:$A-B = \begin{bmatrix} \frac{1}{\pi}(\sin^{-1}(x\pi) + \cos^{-1}(x\pi)) & \frac{1}{\pi}(	an^{-1}(\frac{x}{\pi}) - 	an^{-1}(\frac{x}{\pi})) \ \frac{1}{\pi}(\sin^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi})) & \frac{1}{\pi}(\cot^{-1}(\pi x) + 	an^{-1}(\pi x)) \end{bmatrix}$.Using the identities $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$ and $	an^{-1}(	heta) + \cot^{-1}(	heta) = \frac{\pi}{2}$,we get:$A-B = \begin{bmatrix} \frac{1}{\pi} \cdot \frac{\pi}{2} & 0 \ 0 & \frac{1}{\pi} \cdot \frac{\pi}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2} I$.

Similar Questions

If $A$ and $B$ are square matrices of order $n$ such that $A^{2}-B^{2}=(A-B)(A+B)$,then which of the following will be true?

If $A$ and $B$ are square matrices of order $n \times n$,then ${(A - B)^2}$ is equal to

If $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$,then $A^n = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$ for all $n \in N$. Which of the following is correct?

If $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$,then $A^{100} = $ . . . . . . .

If $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$ and $A^2$ is the identity matrix,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module