Let A = begin{bmatrix} cos theta & -sin theta | vedclass

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

Question

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

Vedclass · Accepted Answer

$\begin{bmatrix} \cos 	heta & -\sin 	heta \ -\sin 	heta & -\cos 	heta \end{bmatrix}$

Vedclass · Answer

(A) Given $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ -\sin 	heta & -\cos 	heta \end{bmatrix}$.First,we find the determinant of $A$:$|A| = (\cos 	heta)(-\cos 	heta) - (-\sin 	heta)(-\sin 	heta) = -\cos^2 	heta - \sin^2 	heta = -(\cos^2 	heta + \sin^2 	heta) = -1$.Next,we find the adjoint of $A$,$	ext{adj}(A)$,by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{adj}(A) = \begin{bmatrix} -\cos 	heta & \sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$:$A^{-1} = \frac{1}{-1} \begin{bmatrix} -\cos 	heta & \sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix} = \begin{bmatrix} \cos 	heta & -\sin 	heta \ -\sin 	heta & -\cos 	heta \end{bmatrix}$.Thus,$A^{-1} = A$.

Let $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{bmatrix}$,then the inverse of $A$ is

Similar Questions

If $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then $A=$

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$,then the matrix $A^{-50}$ when $\theta = \frac{\pi}{12}$ is equal to:

If $A$ is a matrix of order $3$,such that $A(\operatorname{adj} A) = 10I$,then $|\operatorname{adj} A| = $

The inverse of the matrix $\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ is

If $A$ is a symmetric matrix with real entries, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module