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Example: Non-homogeneous consistent system

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

\[\begin{equation*}
  A = \begin{pmatrix}
    1 & -2 & 3 \\
    2 & 1 & -1 \\
    -3 & -4 & 5
  \end{pmatrix}
  \qand
  \vec{b} = \begin{pmatrix}
    3 \\ 9 \\ -15
  \end{pmatrix}.
\end{equation*}
\]
  1. Write the augmented matrix and row-reduce it.
    \[\begin{align*}
      \begin{pmatrix}
        \vphantom{\Big(}
        A & \vec{b} \,
      \end{pmatrix}
      & =
      \begin{pmatrix}
        1 & -2 & 3 & 3 \\
        2 & 1 & -1 & 9 \\
        -3 & -4 & 5 & -15
      \end{pmatrix}
      \\[2ex]
      \operatorname{rref}
      \begin{pmatrix}
        \vphantom{\Big(}
        A & \vec{b} \,
      \end{pmatrix}
      & =
      \begin{pmatrix}
        1 & 0 & \frac{1}{5} & \frac{21}{5} \\
        0 & 1 & -\frac{7}{5} & \frac{3}{5} \\
        0 & 0 & 0 & 0
      \end{pmatrix}
    \end{align*}
    \]
  2. Write the solution in terms of basic variables.
    \[\begin{equation*}
      \begin{aligned}
        x_1 + \frac{1}{5} x_3 & = \frac{21}{5} \\
        x_2 - \frac{7}{5} x_3 & = \frac{3}{5}
      \end{aligned}
      \quad \iff \quad
      \begin{aligned}
        x_1 & = -\frac{1}{5} x_3 + \frac{21}{5} \\
        x_2 & = \frac{7}{5} x_3 + \frac{3}{5}
      \end{aligned}
    \end{equation*}
    \]
  3. Rewrite the solution in the parametric form.
    \[\begin{equation*}
      \vec{x}
      =
      \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}
      =
      \begin{pmatrix}
        -\frac{1}{5} x_3 + \frac{21}{5} \\
        \frac{7}{5} x_3 + \frac{3}{5} \\
        x_3
      \end{pmatrix}
      =
      x_3
      \begin{pmatrix}
        -\frac{1}{5} \\
        \frac{7}{5} \\
        1
      \end{pmatrix}
      +
      \begin{pmatrix}
        \frac{25}{5} \\
        \frac{3}{5} \\
        0
      \end{pmatrix}
      =
      x_3
      \begin{pmatrix} -1 \\ 7 \\ 5 \end{pmatrix}
      +
      \begin{pmatrix} 25 \\ 3 \\ 0 \end{pmatrix}
    \end{equation*}
    \]

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