If A = begin{bmatrix} sin alpha & -cos alpha | vedclass

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

Question

(C) Given,$A = \begin{bmatrix} \sin \alpha & -\cos \alpha \ \cos \alpha & \sin \alpha \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (\si

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(C) Given,$A = \begin{bmatrix} \sin \alpha & -\cos \alpha \ \cos \alpha & \sin \alpha \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (\sin \alpha)(\sin \alpha) - (-\cos \alpha)(\cos \alpha) = \sin^2 \alpha + \cos^2 \alpha = 1$.Next,we find the adjoint of $A$:$	ext{Adj } A = \begin{bmatrix} \sin \alpha & \cos \alpha \ -\cos \alpha & \sin \alpha \end{bmatrix}$.Since $A^{-1} = \frac{1}{|A|} 	ext{Adj } A$,we have:$A^{-1} = \begin{bmatrix} \sin \alpha & \cos \alpha \ -\cos \alpha & \sin \alpha \end{bmatrix}$.Now,calculate $A + A^{-1}$:$A + A^{-1} = \begin{bmatrix} \sin \alpha & -\cos \alpha \ \cos \alpha & \sin \alpha \end{bmatrix} + \begin{bmatrix} \sin \alpha & \cos \alpha \ -\cos \alpha & \sin \alpha \end{bmatrix} = \begin{bmatrix} 2 \sin \alpha & 0 \ 0 & 2 \sin \alpha \end{bmatrix}$.Given $A + A^{-1} = I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,we equate the corresponding elements:$2 \sin \alpha = 1 \implies \sin \alpha = \frac{1}{2}$.Thus,$\alpha = \frac{\pi}{6}$.

If $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$ and $A + A^{-1} = I$,then $\alpha =$

Similar Questions

If $Q$ is the inverse of $A$,where $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $10Q = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3 \end{bmatrix}$,find $x$.

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$,then $A^{-1}$ is given by

If $P$ is a $3 \times 3$ real matrix such that $P^{T} = aP + (a - 1)I$,where $a > 1$,then $..........$

The set of all $2 \times 2$ matrices over the real numbers is not a group under matrix multiplication because

Find the inverse of the matrix $A = \left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$,if it exists.

Vedclass Test Series

Exam Paper Generator

Online Exam Module