MATH 240

Wed. August 28th, 2019


Chap. 1.1 (cont.)/1.2


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.


Echelon Form

Requirements:

Example:


Reduced Echelon Form (REF)

Requirements:

Example:

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.


Reducing a matrix

Step I: Begin with the leftmost nonzero column. This will be a pivot column.
(0366437858391296)\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.
(3912963785803664)\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.
(3912960244203664)\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.
(3912960244200001)\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.
(134320122100001)\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.
(134300122000001)\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.
(102300122000001)\begin{pmatrix} 1&0&-2&3&0\\ 0&1&-2&2&0\\ 0&0&0&0&1 \end{pmatrix}