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If the equations $x+a y-z=0,2 x-y+a z=0$, ax $+y+2 z=0$ have non-trivial solutions, then $a=$
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Verified Answer
The correct answer is:
$-2$
$\left|\begin{array}{ccc}1 & \mathrm{a} & -1 \\ 2 & -1 & \mathrm{a} \\ \mathrm{a} & 1 & 2\end{array}\right|=0$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-$ $\mathrm{aR}_{1}$, we get
$$
\left|\begin{array}{ccc}
1 & a & -1 \\
0 & -1-2 a & a+2 \\
0 & 1-a^{2} & 2+a
\end{array}\right|=0
$$
Expanding along $C_{1}$, we get
$$
(a+2)\left(a^{2}-2 a-2\right)=0 \Rightarrow a=-2,1 \pm \sqrt{3}
$$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-$ $\mathrm{aR}_{1}$, we get
$$
\left|\begin{array}{ccc}
1 & a & -1 \\
0 & -1-2 a & a+2 \\
0 & 1-a^{2} & 2+a
\end{array}\right|=0
$$
Expanding along $C_{1}$, we get
$$
(a+2)\left(a^{2}-2 a-2\right)=0 \Rightarrow a=-2,1 \pm \sqrt{3}
$$
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