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

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

Question

(D) Given $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.The determinant of matrix $A_\alpha$ is $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.The determinant of matrix $A_\alpha$ is $\det(A_\alpha) = \cos^2 \alpha - (-\sin^2 \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1$.We know that for any square matrices $A$ and $B$,$\det(AB) = \det(A) \cdot \det(B)$.Therefore,$\det(A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}) = \det(A_{\pi / 5}) \cdot \det(A_{\pi / 4}) \cdot \det(A_{3 \pi / 10})$.Since $\det(A_\alpha) = 1$ for any value of $\alpha$,we have $\det(A_{\pi / 5}) = 1$,$\det(A_{\pi / 4}) = 1$,and $\det(A_{3 \pi / 10}) = 1$.Thus,$\det(A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}) = 1 	imes 1 	imes 1 = 1$.

If $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$,then the determinant of $A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}$ is:

Similar Questions

If $2\left|\begin{array}{ll}\sin ( A + B ) & \cos ( A + B ) \\ \cos ( A - B ) & \sin ( A - B )\end{array}\right|+\sqrt{3}= 0$,then $A =$ . . . . . . .

The matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is not invertible only if $a =$

Evaluate the determinant: $\left| {\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}} \right|$

If the system of equations $(\alpha + 1)^3 x + (\alpha + 2)^3 y - (\alpha + 3)^3 = 0$,$(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0$,and $x + y - 1 = 0$ is consistent,what is the value of $\alpha$?

$\left|\begin{array}{ccc}\cos 3\pi & \sin 5\pi & \tan 7\pi \\ \sqrt{3} & 1 & 0 \\ \sqrt{5} & 0 & 1\end{array}\right| = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module