Class 12Mathematics3-4. Matrices and DeterminantsExpansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line
EasyThe area of $\triangle PQR$ with the vertices $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$ is $3$ sq. units. Then,$k = $ . . . . . . .
- A$0, 2$
- B$-3, 1$
- C$-1, 3$
- D$1, 3$
Explore More
Similar Questions
$\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5\end{array}\right|=$
If $a, b, c$ are non-zero real numbers and the system of equations $(a - 1)x = y + z,$ $(b - 1)y = z + x,$ $(c - 1)z = x + y$ has a non-trivial solution,then $ab + bc + ca$ equals
DifficultJEE MAIN 2014
View SolutionFind the equation of the line joining $(3, 1)$ and $(9, 3)$ using determinants.
Easy
View SolutionIf $x, y, z$ are not all simultaneously equal to zero,satisfying the system of equations
$(\sin 3 \theta) x - y + z = 0$
$(\cos 2 \theta) x + 4 y + 3 z = 0$
$2 x + 7 y + 7 z = 0$
then the number of principal values of $\theta$ is
$(\sin 3 \theta) x - y + z = 0$
$(\cos 2 \theta) x + 4 y + 3 z = 0$
$2 x + 7 y + 7 z = 0$
then the number of principal values of $\theta$ is
If $k > 1$ and the determinant of the matrix $A^2$, where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$, is $k^2$, then $|\alpha|$ 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