"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 :"

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Accepted Answer

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$$\Delta = \left| {\matrix{ 1 & { - 1} & 1 \cr 2 & 2 & \alpha \cr 3 & { - 1} & 4 \cr } } \right| = 0$$

$\Rightarrow \alpha = 4$

${\Delta _3} = 0$

$$ = \left| {\matrix{ 1 & { - 1} & 5 \cr 2 & 2 & 8 \cr 3 & { - 1} & \beta \cr } } \right| = 0$$

$\Rightarrow \beta = 14$

$\therefore$ ${x^2} - 18x + 56 = 0$

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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 :

Solution

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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 :

Solution

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More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

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 :