JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 2)
For $\alpha, \beta \in \mathbb{R}$, suppose the system of linear equations
$$ \begin{aligned} & x-y+z=5 \\ & 2 x+2 y+\alpha z=8 \\ & 3 x-y+4 z=\beta \end{aligned} $$
has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of :
$$ \begin{aligned} & x-y+z=5 \\ & 2 x+2 y+\alpha z=8 \\ & 3 x-y+4 z=\beta \end{aligned} $$
has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of :
$x^2+18 x+56=0$
$x^2-10 x+16=0$
$x^2+14 x+24=0$
$x^2-18 x+56=0$
Explanation
$$\Delta = \left| {\matrix{ 1 & { - 1} & 1 \cr 2 & 2 & \alpha \cr 3 & { - 1} & 4 \cr } } \right| = 0$$
$$ \Rightarrow \alpha = 4$$
$${\Delta _3} = 0$$
$$ = \left| {\matrix{ 1 & { - 1} & 5 \cr 2 & 2 & 8 \cr 3 & { - 1} & \beta \cr } } \right| = 0$$
$$ \Rightarrow \beta = 14$$
$$\therefore$$ $${x^2} - 18x + 56 = 0$$
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