Ch12. Row Reduction and Echelon Forms

A leading Entry of row : the LEFTMOST nonzero entry.

Echelon Form

  1. All nonzero rows are above any rows of all zeros
  2. Each leading entry of a row is in column to the right of the leading entry of the row above it
  3. The leading entry in each nonzero row is 1
  4. Each leading 1 is the only nonzero entry in its column
\[1 → \left[ \begin{matrix} \bullet & * & * & * \\ 0 & \bullet & * & * \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{matrix} \right] 2 → \left[ \begin{matrix} \bullet & * & * & * \\ 0 & \bullet & * & * \\ 0 & 0 & \bullet & * \\ 0 & 0 & 0 & \bullet \\ \end{matrix} \right]\]

📖 Theorem 1. Uniqueness of the Reduced Echelon Form

Each matrix is row equivalent to one and only one reduced echelon matrix

Row Reduction Algorithm

  1. Begin with the leftmost nonzero column
  2. Select a nonzero entry in the pivot column as a pivot. If necessary, Interchange rows to move this entry into the pivot position.
  3. Row replacement to create zeros in all positions below the pivot.
  4. Apply steps 1-3 to the sub matrix that remain. The combination of steps 1-4 is called forward phase echelon form
  5. Beginning with the rightmost polio and working upward and to the left, created zeros above each pivot. If a pivot is not 1, make it 1 by a scaling operation.
    • Called backward phase reduced echelon form

Solution of linear systems

\(\left[ \begin{matrix} 1 & 0 & -5 & 1 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & 0 & 0 \\ \end{matrix} \right]\)
일 때,
\(x_1 - 5x_3 = 1 \\ x_2 + x_3 = 4 \\ 0 = 0 \\ \left[ \begin{matrix} 6 \\3 \\1 \\ \end{matrix} \right] \left[ \begin{matrix} -9 \\ 6 \\ -2 \\ \end{matrix} \right] \cdots\)
해가 무한대가 되며, General Solution 은 아래와 같다.
\(\begin{cases} x_1 = 1 + 5x_3 & \mbox{basic / leading variables} \\ x_2 = 4 - x_3 & \mbox{basic / leading variables} \\ x_3 \mbox{ is free} & \mbox{free variable} \end{cases}\)

📖 Theorem 2. Existence and Uniqueness Theorem

A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column - that is, if and only if and echelon form of the augmented matrix has no row of the forms.
\(\left[ \begin{matrix} 0 & \cdots & 0 & b \end{matrix} \right]\) with \(b\) is nonzero.
If a linear system is consistent, then the solution set contains either (i) a unique solution, when there are no free variables, or (ii) infinitely many solutions, when there is at least one free variables.