Determine if the system has a nontrivial solution. Try to use as few row operati

Determine if the system has a nontrivial solution. Try to use as few row operations as possible.
$2{x}_{1}-5{x}_{2}+8{x}_{3}=0$
$-2{x}_{1}-7{x}_{2}+{x}_{3}=0$
$4{x}_{1}+2{x}_{2}+7{x}_{3}=0$
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Step 1
The given system is
$2{x}_{1}-5{x}_{2}+8{x}_{3}=0$
$-2{x}_{1}-7{x}_{2}+{x}_{3}=0$
$4{x}_{1}+2{x}_{2}+7{x}_{3}=0$
The augmented matrix is
$\left[\begin{array}{cccc}2& -5& 8& 0\\ -2& -7& 1& 0\\ 4& 2& 7& 0\end{array}\right]$

$\left[\begin{array}{cccc}4& 2& 7& 0\\ 0& -6& \frac{9}{2}& 0\\ 0& -6& \frac{9}{2}& 0\end{array}\right]\begin{array}{c}{R}_{3}-{R}_{2}\to {R}_{3}\\ -1/6{R}_{2}\to {R}_{2}\end{array}\text{~}\left[\begin{array}{cccc}4& 2& 7& 0\\ 0& 1& -\frac{3}{4}& 0\\ 0& 0& 0& 0\end{array}\right]$

Thus we get
${x}_{1}+\frac{17}{8}{x}_{3}=0⇒{x}_{1}=-\frac{17}{8}{x}_{3}$
${x}_{2}-\frac{3}{4}{x}_{3}=0⇒{x}_{2}=\frac{3}{4}{x}_{3}$
Hence, the given system has a nontrivial solution because variable ${x}_{3}$ is a free.
Result
The system has a nontrivial solution because variable ${x}_{3}$ is a free