# Write the solution of the given homogeneous system in parametric

Write the solution of the given homogeneous system in parametric vector form
$$\displaystyle{3}{x}_{{1}}+{3}{x}_{{2}}+{6}{x}_{{3}}={0}$$
$$\displaystyle-{9}{x}_{{1}}-{9}{x}_{{2}}-{18}{x}_{{3}}={0}$$
$$\displaystyle-{6}{x}_{{2}}-{6}{x}_{{3}}={0}$$
where the solution set is $x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$

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To describe the system of equation in the parametric vector form, solve for $$\displaystyle{x}_{{1}},\ {x}_{{2}}$$ and $$\displaystyle{x}_{{3}}$$.
$$\displaystyle{3}{x}_{{1}}+{3}{x}_{{2}}+{6}{x}_{{3}}={0}$$
$$\displaystyle-{9}{x}_{{1}}-{9}{x}_{{2}}-{18}{x}_{{3}}={0}$$
$$\displaystyle-{6}{x}_{{2}}-{6}{x}_{{3}}={0}$$
Let Solving equation we get:
$$\displaystyle-{6}{x}_{{2}}={6}{x}_{{3}}$$
$$\displaystyle{x}_{{2}}=-{x}_{{3}}$$
Substitute $$\displaystyle{x}_{{2}}=-{x}_{{3}}$$ in equation, we get:
$$\displaystyle{3}{x}_{{1}}-{3}{x}_{{3}}+{6}{x}_{{3}}={0}$$
$$\displaystyle{3}{x}_{{1}}=-{3}{x}_{{3}}$$
$$\displaystyle{x}_{{1}}=-{x}_{{3}}$$
Thus, the paranetric vector form for system of equation is as follows:
$x=\begin{bmatrix}-1\\-1\\1\end{bmatrix}$