If $\alpha, \beta, \gamma(\alpha \lt \beta \lt \gamma)$ are the values of $x$ such that $\left[\begin{array}{ccc}x-2 & 0 & 1 \ 1 & x+3 & 2 \ 2 & 0 & 2 x-1\end{array} ight]$ is a singular matrix then $2 \alpha+3 \beta+4 \gamma=$

Question

If $\alpha, \beta, \gamma(\alpha \lt \beta \lt \gamma)$ are the values of $x$ such that $\left[\begin{array}{ccc}x-2 & 0 & 1 \ 1 & x+3 & 2 \ 2 & 0 & 2 x-1\end{array}ight]$ is a singular matrix then $2 \alpha+3 \beta+4 \gamma=$

Aicharya · Accepted Answer

$\begin{aligned} & \quad\left|\begin{array}{ccc}x-2 & 0 & 1 \ 1 & x+3 & 2 \ 2 & 0 & 2 x-1\end{array}ight|=0 \ & \Rightarrow(x-2)[(x+3)(2 x-1)]-2(x+3)=0 \ & \Rightarrow x(x+3)(2 x-5)=0 	herefore x=0,-3, \frac{5}{2} \ & 	ext { So, } \alpha=-3, \beta=0, \gamma=\frac{5}{2} \Rightarrow 2 \alpha+3 \beta+4 \gamma=4\end{aligned}$

If $\alpha, \beta, \gamma(\alpha \lt \beta \lt \gamma)$ are the values of $x$ such that…

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