For , R, suppose the system of linear equations [ aligned & x-y+z=5 & 2 x+2 y+ z=8 & 3 x-y…

Question

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 :

Accepted Answer

$$\Delta  = \left| {\begin{pmatrix} 1 & { - 1} & 1  \
   2 & 2 & \alpha   \
   3 & { - 1} & 4  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow \alpha  = 4$$

$${\Delta _3} = 0$$

$$ = \left| {\begin{pmatrix} 1 & { - 1} & 5  \
   2 & 2 & 8  \
   3 & { - 1} & \beta   \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow \beta  = 14$$

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

Solutions · 1