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

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

Question

(B) Given $A = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.We know that for any square matrix $A$,$A \cdot

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $A = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.We know that for any square matrix $A$,$A \cdot 	ext{adj}(A) = |A| I$,where $I$ is the identity matrix.First,calculate the determinant of $A$:$|A| = (\cos \alpha)(\cos \alpha) - (\sin \alpha)(-\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1$.Thus,$A \cdot 	ext{adj}(A) = 1 \cdot \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Comparing this with the given matrix $\begin{bmatrix} k & 0 \ 0 & k \end{bmatrix}$,we get $k = 1$.

If $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$ and $A \cdot \text{adj}(A) = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$,then $k$ is equal to

Similar Questions

If $A$ is a non-singular matrix and $A^2 - A + I = 0$, then $A^{-1} = \dots$

If $A = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $\operatorname{adj} A = $

By using elementary operations,find the inverse of the matrix $A=\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]$.

Let $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$. If $A^{-1} = \alpha I + \beta A$,where $\alpha, \beta \in \mathbb{R}$ and $I$ is a $2 \times 2$ identity matrix,then $4(\alpha - \beta)$ is equal to:

Let $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module