If $\left| {{\kern 1pt} \begin{array}{*{20}{c}}1&2&3\\2&x&3\\3&4&5\end{array}\,} \right| = 0,$ then  $x =$

  • A

    $-5/2$

  • B

    $-2/5$

  • C

    $5/2$

  • D

    $2/5$

Similar Questions

If the following system of linear equations

$2 x+y+z=5$

$x-y+z=3$

$x+y+a z=b$

has no solution, then :

  • [JEE MAIN 2021]

If $-9 $ is a root of the equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&2\\7&6&x\end{array}\,} \right| = 0$ then the other two roots are

  • [IIT 1983]

Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in R$. If $x(\alpha, 1,2)+y(1, \beta, 2)$ $+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in R, x y z \neq$ 0 , then $6 \alpha+4 \beta+\gamma$ is equal to..............

  • [JEE MAIN 2024]

Evaluate $\Delta=\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|$

If $\left| {\begin{array}{*{20}{c}}
{a - b - c}&{2a}&{2a}\\
{2b}&{b - c - a}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|$ $ = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}$ , $x   \ne 0$ and $a + b + c \ne 0$, then $x$ is equal to

  • [JEE MAIN 2019]