MATH 240

Wed. August 28th, 2019

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Chap. 1.1 (cont.)/1.2

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Elementary Row Operations

1. Interchange two rows
2. Add a constant multiple of one row to another row
3. Multiply a row by a nonzero constant

Definition: Two matrices A and B are row-equivalent if B can be created by only performing elementary row operations on A.
Row-equivalency follows the transitive property, just like equality does.

Theorem: If two systems of linear equations have row-equivalent augmented matrices, then they have the same solution set.

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Echelon Form

Requirements:

• Rows with all zeroes are below rows with nonzero entries.
• Leading entry of each row is in a column to the right of the leading entry of the row above it.
  • (Leading entry = first nonzero entry of a row)
• All entries in a column below a leading entry are zeroes.

Example:

• $\begin{pmatrix} 0&1&0\\1&1&0\\0&0&1 \end{pmatrix}$ IS NOT in echelon form.
• $\begin{pmatrix} 1&1&0\\0&1&0\\0&0&1 \end{pmatrix}$ IS in echelon form. (Only differs by one row operation from previous matrix)

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Reduced Echelon Form (REF)

Requirements:

• Must be in echelon form.
• All leading entries must be 1.
• Each leading entry must be the only nonzero entry in its row/column.

Example:

• $\begin{pmatrix} 1&1\\0&2 \end{pmatrix}$ IS NOT in REF.
• $\begin{pmatrix} 1&1\\0&1 \end{pmatrix}$ IS NOT in REF.
• $\begin{pmatrix} 1&0\\0&1 \end{pmatrix}$ IS in REF. (Created from original matrix using only row operations, so they are row equivalent)

Theorem: Every matrix A is row-equivalent to exactly ONE reduced echelon matrix.
This also means that the order that the row operations are performed in to create the REF matrix doesn't matter in the end.

Definition: A pivot position in a matrix A is a location that corresponds to a leading entry in the reduced echelon form of A. A pivot column of A is a column that contains a pivot position.

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Reducing a matrix

Step I: Begin with the leftmost nonzero column. This will be a pivot column.
$\begin{pmatrix} 0&3&-6&6&4\\ 3&-7&8&-5&8\\ 3&-9&12&-9&6 \end{pmatrix}$

Step II: Interchange rows so that zeroes are at the bottom, if necessary.
$\begin{pmatrix} 3&-9&12&-9&6\\ 3&-7&8&-5&8\\ 0&3&-6&6&4 \end{pmatrix}$

Step III: Use row operations to create zeroes in all positions below the pivot.
$\begin{pmatrix} 3&-9&12&-9&6\\ 0&2&-4&4&2\\ 0&3&-6&6&4 \end{pmatrix}$

Step IV: Ignore the row containing the pivot in the current column (and any rows above it). Repeat steps I-IV to get an echelon matrix.
$\begin{pmatrix} 3&-9&12&-9&6\\ 0&2&-4&4&2\\ 0&0&0&0&1 \end{pmatrix}$

Step V: Multiply each row by the reciprocal of its leading entry.
$\begin{pmatrix} 1&-3&4&-3&2\\ 0&1&-2&2&1\\ 0&0&0&0&1 \end{pmatrix}$

Step VI: Now start at the rightmost pivot and use its row to make the entries above it zeroes.
$\begin{pmatrix} 1&-3&4&-3&0\\ 0&1&-2&2&0\\ 0&0&0&0&1 \end{pmatrix}$

Step VII: Repeat step VI for each pivot.
$\begin{pmatrix} 1&0&-2&3&0\\ 0&1&-2&2&0\\ 0&0&0&0&1 \end{pmatrix}$