If $F(\alpha ) = \begin{bmatrix} \cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $G(\beta ) = \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta \end{bmatrix}$,then $[F(\alpha ) G(\beta )]^{-1} = $
- A$F(\alpha ) - G(\beta )$
- B$- F(\alpha ) - G(\beta )$
- C$[F(\alpha )]^{-1} [G(\beta )]^{-1}$
- D$[G(\beta )]^{-1} [F(\alpha )]^{-1}$
Explore More
Similar Questions
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
Medium
View SolutionIf $\operatorname{adj} B = A$ and $|P| = |Q| = 1$, then $\operatorname{adj}(Q^{-1} B P^{-1}) = $
MediumWBJEE 2025
View SolutionIf $\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = I_2$,then $A =$
If $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$,then verify that $A \text{ adj } A = |A| I$. Also,find $A^{-1}$.
Medium
View SolutionFind the inverse of the matrix,if it exists: $\left[\begin{array}{ll}3 & 1 \\ 5 & 2\end{array}\right]$
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo