# Write the solution set of the given homogeneous system in

Write the solution set of the given homogeneous system in parametric vector form.
$$\displaystyle{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={0}$$
$$\displaystyle{x}_{{1}}+{4}{x}_{{2}}-{8}{x}_{{3}}={0}$$
$$\displaystyle-{3}{x}_{{1}}-{7}{x}_{{2}}+{9}{x}_{{3}}={0}$$

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The given system is
$$\displaystyle{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={0}$$
$$\displaystyle{x}_{{1}}+{4}{x}_{{2}}-{8}{x}_{{3}}={0}$$
$$\displaystyle-{3}{x}_{{1}}-{7}{x}_{{2}}+{9}{x}_{{3}}={0}$$
The augmented matrix is
$\begin{bmatrix}1&3&-5&0\\1&4&-8&0\\-3&-7&9&0\end{bmatrix}$
Reduce matrix to row echelon form
$\begin{bmatrix}-3&-7&9&0\\0&\frac{5}{3}&-5&0\\0&\frac{2}{3}&-2&0\end{bmatrix}\begin{matrix}R_3-2/5R_2\to R_3\\3/5R_2\to R_2 \end{matrix}\sim\begin{bmatrix}-3&-7&9&0\\0&1&-3&0\\0&0&0&0 \end{bmatrix}$
$\begin{bmatrix}-3&-7&9&0\\0&1&-3&0\\0&0&0&0 \end{bmatrix}\begin{matrix}R_1+7R_2\to R_1\\-1/3R_1\to R_1\end{matrix}\begin{bmatrix}1&0&4&0\\0&1&-3&0\\0&0&0&0\end{bmatrix}$
Thus we get
$$\displaystyle{x}_{{1}}+{4}{x}_{{3}}={0}\Rightarrow{x}_{{1}}=-{4}{x}_{{3}}$$
$$\displaystyle{x}_{{2}}-{3}{x}_{{3}}={0}\Rightarrow{x}_{{2}}={3}{x}_{{3}}$$
Hence, the given homogeneous system in parametric vector form is
$x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=x_3\begin{bmatrix}-4\\3\\1\end{bmatrix}$