If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$,then the value of the determinant of $A^{-1}$ is
- A$-6$
- B$-\frac{1}{6}$
- C$\frac{1}{36}$
- D$36$
Explore More
Similar Questions
The inverse of matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is
Easy
View SolutionIf $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$,such that $A^{2} - 4A + 3I = 0$,then $A^{-1} =$
If $B$ is the inverse of a third order matrix $A$ and $\det B = k$,then $(\operatorname{adj}(\operatorname{adj} A))^{-1} =$
If $A$ is a symmetric matrix with real entries, then
If $A = \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix}$,then $\text{adj}(3A^2 + 12A)$ is equal to
Vedclass 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