If A(theta)=begin{bmatrix} i sin theta & cos | vedclass

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

Question

(C) Given $A(	heta)=\begin{bmatrix} i \sin 	heta & \cos 	heta \ \cos 	heta & i \sin 	heta \end{bmatrix}$.The determinant is $\operatorname{det}

Vedclass · Accepted Answer

$\operatorname{det}[A(	heta)]^{-1}=1$

Vedclass · Answer

(C) Given $A(	heta)=\begin{bmatrix} i \sin 	heta & \cos 	heta \ \cos 	heta & i \sin 	heta \end{bmatrix}$.The determinant is $\operatorname{det} A(	heta) = (i \sin 	heta)(i \sin 	heta) - (\cos 	heta)(\cos 	heta) = i^2 \sin^2 	heta - \cos^2 	heta$.Since $i^2 = -1$,we have $\operatorname{det} A(	heta) = -\sin^2 	heta - \cos^2 	heta = -(\sin^2 	heta + \cos^2 	heta) = -1$.Since the determinant is a constant $-1$ regardless of $	heta$,we have $\operatorname{det} A(\pi+	heta) = -1$,$\operatorname{det} A(-	heta) = -1$,and $\operatorname{det} A(	heta) = -1$.Checking the options:Option $A$: $\operatorname{det} A(\pi+	heta) = -1$ and $\operatorname{det} A(-	heta) = -1$. So,$-1 = -1$ (True).Option $B$: $\operatorname{det} A(-	heta) = -1$ and $\operatorname{det} A(	heta) = -1$. So,$-1 = -1$ (True).Option $C$: $\operatorname{det}[A(	heta)]^{-1} = \frac{1}{\operatorname{det} A(	heta)} = \frac{1}{-1} = -1$. The statement says $1$,which is False.Option $D$: $\operatorname{det} A(-	heta) = -1$ (True).Therefore,the statement that is not true is $C$.

If $A(\theta)=\begin{bmatrix} i \sin \theta & \cos \theta \\ \cos \theta & i \sin \theta \end{bmatrix}$ is a matrix,where $i=\sqrt{-1}$,then which of the following is not true?

Similar Questions

Let $A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$. If $\det(B) = 66$,then $\det(\text{adj}(A))$ equals:

The number of elements in the set $\{A=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} : a, b, d \in \{-1, 0, 1\} \text{ and } (I-A)^3 = I-A^3 \}$,where $I$ is the $2 \times 2$ identity matrix,is:

The value of $\left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ is

If $A$ is a square matrix and $A^2+I=2 A$,then $A^9=$

Let $P$ be a square matrix such that $P^2 = I - P$. For $\alpha, \beta, \gamma, \delta \in N$,if $P^\alpha + P^\beta = \gamma I - 29 P$ and $P^\alpha - P^\beta = \delta I - 13 P$,then $\alpha + \beta + \gamma - \delta$ is equal to $........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module