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3 and 4 .Determinants and Matrices
hard
The sum of the real roots of the equation $\left| {\begin{array}{*{20}{c}}
x&{ - 6}&{ - 1}\\
2&{ - 3x}&{x - 3}\\
{ - 3}&{2x}&{x = 2}
\end{array}} \right| = 0$ is equal to
A
$-4$
B
$0$
C
$6$
D
$1$
(JEE MAIN-2019)
Solution
By expansion, we get
$ – 5{x^3} + 30x – 30 + 5x = 0$
$ \Rightarrow – 5{x^3} + 35x – 30 = 0$
$ \Rightarrow {x^3} – 7x + 6 = 0$, All roots srea real
So, sum of roots $=0$
Standard 12
Mathematics
Similar Questions
Let $\alpha, \beta$ and $\gamma$ be real numbers. consider the following system of linear equations
$x+2 y+z=7$
$x+\alpha z=11$
$2 x-3 y+\beta z=\gamma$
Match each entry in List – $I$ to the correct entries in List-$II$
| List – $I$ | List – $II$ |
| ($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has | ($1$) a unique solution |
| ($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has | ($2$) no solution |
|
($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has |
($3$) infinitely many solutions |
| ($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has | ($4$) $x=11, y=-2$ and $z=0$ as a solution |
| ($5$) $x=-15, y=4$ and $z=0$ as a solution |
Then the system has
