A=left[begin{array}{rr}cos theta & -sin theta | vedclass

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

Question

(B) Given,$A=\left[\begin{array}{rr}\cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta\end{array}ight]$.Since $AB=BA=I$,$B$ is the inverse of $

Vedclass · Accepted Answer

$\left[\begin{array}{rr}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$

Vedclass · Answer

(B) Given,$A=\left[\begin{array}{rr}\cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta\end{array}ight]$.Since $AB=BA=I$,$B$ is the inverse of $A$,i.e.,$B=A^{-1}$.The determinant of $A$ is $|A| = \cos^2 	heta - (-\sin^2 	heta) = \cos^2 	heta + \sin^2 	heta = 1$.The inverse of a $2 	imes 2$ matrix $\left[\begin{array}{rr}a & b \ c & d\end{array}ight]$ is $\frac{1}{ad-bc} \left[\begin{array}{rr}d & -b \ -c & a\end{array}ight]$.Applying this to $A$,we get $B = A^{-1} = \frac{1}{1} \left[\begin{array}{rr}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$.Thus,$B = \left[\begin{array}{rr}\cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta\end{array}ight]$.

$A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$ and $AB=BA=I$,then $B$ is equal to

Similar Questions

If $A = \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{bmatrix}$ and $A(\operatorname{adj} A) = K I$,then the value of $K$ is (where $I$ is the unit matrix of order $3$).

If $A = \begin{bmatrix} 0 & 3 \\ 2 & 0 \end{bmatrix}$ and $A^{-1} = \lambda (adj(A))$,then $\lambda = $

If $X$ is a square matrix of order $3 \times 3$ and $\lambda$ is a scalar,then $adj(\lambda X)$ is equal to

If $A$ is a square matrix of order $n$,$|A| = D$,and $|adj(A)| = D'$,then:

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 1 & 6 \end{bmatrix}$ and $|\text{adj}(\text{adj } A)|(\text{adj } A)^{-1} = kA$,then $k = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module