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

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

Question

(D) 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

(D) 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 $A$ 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}$.Finally,the inverse 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,$A^{-1} = A$.

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

Similar Questions

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$,then the value of the determinant of $A^{-1}$ is

If $B = \begin{bmatrix} 3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A| = 4$,then $\alpha$ is equal to:

Let $A = (a_{ij})_{1 \leq i, j \leq 3}$ be a $3 \times 3$ invertible matrix where each $a_{ij}$ is a real number. Denote the inverse of the matrix $A$ by $A^{-1}$. If $\sum_{j=1}^3 a_{ij} = 1$ for $1 \leq i \leq 3$,then:

If $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$, then the matrix $A^{-3}$ when $\theta = \pi/6$ is equal to...

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, and $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, then $|\alpha \cdot \beta \cdot \gamma| = $ (where $| \cdot |$ denotes the absolute value)

Vedclass Test Series

Exam Paper Generator

Online Exam Module