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3.2 Reduced Row Echelon Form

Row Echelon Form

row echelon: reduce by row operations (Gaussian elimination). All numbers below a pivot (leftmost non-zero in a row) are zero.

\[ \begin{bmatrix} 1 & 3 & 9 \\ 2 & 1 & 8 \end{bmatrix} \xrightarrow{(-2)R_1 + R_2} \begin{bmatrix} 1 & 3 & 9 \\ 0 & -5 & -10 \end{bmatrix} \]

pivots: The numbers 1 and -5 in the resulting matrix. #'s below are zero.

Reduced Row Echelon Form

reduced row echelon: row echelon AND every pivot is the only non-zero # in its column and every pivot is 1.

\[ \begin{bmatrix} 1 & 3 & 9 \\ 0 & -5 & -10 \end{bmatrix} \]\[ \xrightarrow{(-\frac{1}{5})R_2} \begin{bmatrix} 1 & 3 & 9 \\ 0 & 1 & 2 \end{bmatrix} \xrightarrow{(-3)R_2 + R_1} \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \end{bmatrix} \]

reduced row echelon form (RREF): The final matrix where pivots are 1 and are the only non-zero entries in their columns.

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Uniqueness of Row Echelon Form

row echelon for a matrix is NOT unique.

\[ \begin{bmatrix} 1 & 3 & 9 \\ 2 & 1 & 8 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 3 & 9 \\ 0 & -5 & -10 \end{bmatrix} \text{ (one row echelon)} \]

This corresponds to the system:
\( x_1 + 3x_2 = 9 \)
\( 2x_1 + x_2 = 8 \)

Another example:

\[ \text{another: } \begin{bmatrix} 2 & 1 & 8 \\ 1 & 3 & 9 \end{bmatrix} \xrightarrow{(-\frac{1}{2})R_1 + R_2} \begin{bmatrix} 2 & 1 & 8 \\ 0 & 5/2 & 5 \end{bmatrix} \text{ (another row echelon)} \]

Geometric Interpretation

Solution is the same! Row reduction \( \rightarrow \) replacing a system w/ another system with same solution.

Coordinate graph with two intersecting lines (red and green) in the x_1, x_2 plane.
Coordinate graph showing a modified system of lines intersecting at the same point.
RREF\( \rightarrow \)
Coordinate graph in RREF showing one horizontal and one vertical line intersecting at the solution.
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Row Echelon and Reduced Row Echelon Form

Row echelon: change lines but keep intersection (infinitely many possibilities)

rref: vertical and horizontal lines only (only one possibility)

rref for a matrix is unique

\[ \begin{bmatrix} 1 & 3 & 9 \\ 2 & 1 & 8 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 3 & 9 \\ 0 & -5 & -10 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \end{bmatrix} \]

Alternative path:

\[ \begin{bmatrix} 2 & 1 & 8 \\ 0 & 5/2 & 5 \end{bmatrix} \rightarrow \begin{bmatrix} 2 & 1 & 8 \\ 0 & 1 & 2 \end{bmatrix} \rightarrow \begin{bmatrix} 2 & 0 & 6 \\ 0 & 1 & 2 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \end{bmatrix} \]

Note: Both paths lead to the same unique Reduced Row Echelon Form (rref).

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Solving Systems with rref

rref → no back substitution necessary (read rows for solution)

Example

\[ \begin{cases} x_1 + 3x_2 + 2x_3 = 12 \\ 2x_1 + 5x_2 + 2x_3 = 12 \\ 2x_1 + 7x_2 + 7x_3 = 43 \end{cases} \]
\[ \begin{bmatrix} 1 & 3 & 2 & 12 \\ 2 & 5 & 2 & 12 \\ 2 & 7 & 7 & 43 \end{bmatrix} \]

make it rref

Gauss-Jordan elimination

Step 1: Eliminate first column

\[ (-2)R_1 + R_2 \]\[ (-2)R_1 + R_3 \]
\[ \rightarrow \begin{bmatrix} 1 & 3 & 2 & 12 \\ 0 & -1 & -2 & -12 \\ 0 & 1 & 3 & 19 \end{bmatrix} \]

Step 2: Eliminate second column

\[ R_2 + R_3 \]
\[ \rightarrow \begin{bmatrix} 1 & 3 & 2 & 12 \\ 0 & -1 & -2 & -12 \\ 0 & 0 & 1 & 7 \end{bmatrix} \]
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\((3)R_2 + R_1 \rightarrow\)
\[\begin{bmatrix} 1 & 0 & -4 & -24 \\ 0 & -1 & -2 & -12 \\ 0 & 0 & 1 & 7 \end{bmatrix}\]
\(\begin{matrix} (2)R_3 + R_2 \\ (4)R_3 + R_1 \end{matrix} \rightarrow\)
\[\begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & -1 & 0 & 2 \\ 0 & 0 & 1 & 7 \end{bmatrix}\]
\(\xrightarrow{(-1)R_2}\)
\[\begin{matrix} x_1 & x_2 & x_3 \\ \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 7 \end{bmatrix} \end{matrix}\]

row 3: \(x_3 = 7\)

row 2: \(x_2 = -2\)

row 1: \(x_1 = 4\)

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Example: Homogeneous System

\[\begin{aligned} x_1 + 3x_2 - 15x_3 + 7x_4 &= 0 \\ x_1 + 4x_2 - 19x_3 + 10x_4 &= 0 \\ 2x_1 + 5x_2 - 26x_3 + 11x_4 &= 0 \end{aligned}\]

right side all zeros

homogeneous system

\(\uparrow\) has nothing to do with \(v = \frac{y}{x}\)

\[\begin{bmatrix} 1 & 3 & -15 & 7 & 0 \\ 1 & 4 & -19 & 10 & 0 \\ 2 & 5 & -26 & 11 & 0 \end{bmatrix}\]
\(\begin{matrix} (-1)R_1 + R_2 \\ (-2)R_1 + R_3 \end{matrix} \rightarrow\)
\[\begin{bmatrix} 1 & 3 & -15 & 7 & 0 \\ 0 & 1 & -4 & 3 & 0 \\ 0 & -1 & 4 & -3 & 0 \end{bmatrix}\]
\(\begin{matrix} (-3)R_2 + R_1 \\ R_2 + R_3 \end{matrix} \rightarrow\)
\[\begin{matrix} x_1 & x_2 & x_3 & x_4 \\ \begin{bmatrix} 1 & 0 & -3 & -2 & 0 \\ 0 & 1 & -4 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{matrix}\]
RREF
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Linear Systems: Infinitely Many Solutions

row 3: \( 0 = 0 \)

infinitely-many solutions

\( \rightarrow \) free variable(s)

variables w/ no pivots in their columns are free

here, \( x_3 \) and \( x_4 \) are free

\( x_4 = r \)

\( x_3 = t \)

\( x_2 - 4x_3 + 3x_4 = 0 \quad \rightarrow \quad x_2 = 4t - 3r \)

\( x_1 - 3x_3 - 2x_4 = 0 \quad \rightarrow \quad x_1 = 3t + 2r \)