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I have just started doing linear algebra and have a somewhat basic question that all other answers I've seen on here are a bit over my head for. I am wondering what the practical differences between row echelon and reduced row echelon. Specifically so I know I have matrices in the correct forms, and am using the correct terminology when writing answers.

These definitions and terms seem to be correct, but I was wondering if anyone could verify that it is:

Rules for elimination

Each of the forms of elimination can only have one of 3 operations per step:

  1. Subtract or add a row to/from another row
  2. Multiply a scalar by a row and subtract or add it from another row
  3. Multiply/Divide a row by a scalar

Row Echelon

This is achieved with Gaussian Elimination, and requires that the matrix has the following properties:

  1. Each row going down the matrix must have a 1 in the leftmost position
  2. Each row must have the leftmost coefficient at least 1 column to the right of the row above it

For example:

$$ \begin{bmatrix} 1 & 3 & 2 & 0\\ 0 & 1 & 3 & 2\\ 0 & 0 & 1 & -4 \\ \end{bmatrix} $$

Reduced row Echelon

The same requirements as row echelon, except now you use Gauss-Jordan Elimination, and there is an additional requirement that:

  1. Each of the leftmost coefficients must be the only non-zero value in a given column

For example

$$ \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & -4 \\ \end{bmatrix} $$

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Your summaries of 'Row echelon' and 'Reduced row echelon' are completely correct, but there is a slight issue with the rules for elimination. Typically, these are given as

(1) Interchange rows;

(2) Multiply a row by a non-zero scalar; and

(3) Add a scalar multiple of one row to another row.

Note that the third case covers subtraction of rows since you just consider this as multiplying one row by a negative scalar and adding to another row.

Now in your case, you are missing the 'interchange rows' rule. Further, your 1. and 3. considered together imply 2., so one of your listed rules (rule 2.) is redundant.

As for practical differences between REF and RREF, one practical difference is that in some situations it may be quicker or more computationally efficient to just compute REF. For example, solving a system of linear equations, it is typically quicker to just compute the REF of a system, and then solve the system by 'back substitution,' rather than spending more time computing the RREF of the system.

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