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
EasyIf $\alpha, \beta, \gamma$ $(\alpha < \beta < \gamma)$ are the values of $x$ such that $\begin{vmatrix} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2x-1 \end{vmatrix} = 0$ is a singular matrix,then $2\alpha + 3\beta + 4\gamma = $
- A$4$
- B$0$
- C$1$
- D$2$
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If ${\left| {\begin{array}{cc} 4 & 1 \\ 2 & 1 \end{array}} \right|^2} = \left| {\begin{array}{cc} 3 & 2 \\ 1 & x \end{array}} \right| - \left| {\begin{array}{cc} x & 3 \\ -2 & 1 \end{array}} \right|$,then $x =$
Medium
View SolutionThe value of $a$ for which the system of equations has a non-zero solution is
$a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$
$a x+(a+1) y+(a+2) z=0$
$x+y+z=0$
$a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$
$a x+(a+1) y+(a+2) z=0$
$x+y+z=0$
If one of the roots of $\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0$ is $-10$,then the other roots are
If $\left|\begin{array}{lll}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x, y, z$ are all distinct,then $x y z=$
If the system of equations $ax + y + z = 0$,$x + by + z = 0$ and $x + y + cz = 0$,where $a, b, c \neq 1$,has a non-trivial solution,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ is
Medium
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