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Example: Homogeneous consistent system

Solve \(A \vec{x} = \vec{0}\), given

\[\begin{equation*}
  A = \begin{pmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ -3 & -4 & 5 \end{pmatrix}.
\end{equation*}
\]

1. Write the augmented matrix and row-reduce it.

   \[\begin{align*}
     \begin{pmatrix}
       \vphantom{\Big(}
       A & \vec{0} \,
     \end{pmatrix}
     & =
     \begin{pmatrix}
       1 & -2 & 3 & 0 \\
       2 & 1 & -1 & 0 \\
       -3 & -4 & 5 & 0
     \end{pmatrix}
     \\[2ex]
     \operatorname{rref}
     \begin{pmatrix}
       \vphantom{\Big(}
       A & \vec{0} \,
     \end{pmatrix}
     & =
     \begin{pmatrix}
       1 & 0 & \frac{1}{5} & 0 \\
       0 & 1 & -\frac{7}{5} & 0 \\
       0 & 0 & 0 & 0
     \end{pmatrix}
   \end{align*}
   \]

2. Write the solution in terms of basic variables.

   \[\begin{equation*}
     \begin{aligned}
       x_1 + \frac{1}{5} x_3 & = 0 \\
       x_2 - \frac{7}{5} x_3 & = 0
     \end{aligned}
     \quad \iff \quad
     \begin{aligned}
       x_1 & = -\frac{1}{5} x_3 \\
       x_2 & = \frac{7}{5} x_3
     \end{aligned}
   \end{equation*}
   \]

3. Rewrite the solution in the parametric form.

   \[\begin{equation*}
     \vec{x}
     =
     \begin{pmatrix}
       x_1 \\ x_2 \\ x_3
     \end{pmatrix}
     =
     \begin{pmatrix}
       -\frac{1}{5} x_3 \\ \frac{7}{5} x_3 \\ x_3
     \end{pmatrix}
     = x_3
     \begin{pmatrix}
       -\frac{1}{5} \\ \frac{7}{5} \\ 1
     \end{pmatrix}
     = x_3
     \begin{pmatrix}
       -1 \\ 7 \\ 5
     \end{pmatrix}.
   \end{equation*}
   \]

   Note. This is a line in \(\mathbb{R}^3\).