Class 12Mathematics3 and 4 .Determinants and MatricesExpansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line
EasyThe solutions of the equation $\left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| = 0$ are:
- A$x = 1, 2$
- B$x = 2, 3$
- C$x = 1, p, 2$
- D$x = 1, 2, -p$
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Similar Questions
The number of positive integral solutions of the equation $\left| \begin{array}{ccc} 1 - \lambda & 2 & 1 \\ -3 & \lambda & -2 \\ 2 & -2 & 1 + \lambda \end{array} \right| = 0$ is
Medium
View SolutionEvaluate the determinant: $\left|\begin{array}{ccc}1 & a & b \\ 1 & a+b & b \\ 1 & a & a+b\end{array}\right| = $ . . . . . . .
The value of the determinant given below $\left| \begin{matrix} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{matrix} \right|$ is
Difficult
View SolutionIf the system of equations
$x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0$
$x+(\cos \alpha) y+(\sin \alpha) z=0$
$x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution,then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to :
$x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0$
$x+(\cos \alpha) y+(\sin \alpha) z=0$
$x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution,then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to :
DifficultJEE MAIN 2024
View SolutionThe area of a triangle is $5$. If two of its vertices are $(2, 1)$ and $(3, -2)$ and the third vertex lies on the line $y = x + 3$,then the third vertex is
MediumIIT 1978
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