[Math] Reduced Row vs. Reduced Row-Echelon

calculuslinear algebramatrices

I'm a little confused as to what the difference between reduced row and reduced row-echelon is. I have this following problem:

Row reduce the following matrix in as few steps as possible.
\begin{bmatrix}3&-1&-1&-1&0&0&0\\-1&1&0&0&0&0&0\\-1&0&1&0&0&0&0\\-1&0&0&1&0&0&0\\0&0&0&0&2&1&1\\0&0&0&0&1&2&1\\0&0&0&0&1&1&2\end{bmatrix}

I figure using simple row operations would reduce it, but I'm confused as to which method I should use. Any help?

Best Answer

There is a difference between row-echelon form, and reduced row-echelon form..

Specifically, a matrix is in row echelon form if:

  • all nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix), and

  • the leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it (some texts add the condition that the leading coefficient must be 1).

These two conditions imply that all entries in a column below a leading coefficient are zeros. [2]

$$ $$

A matrix is in reduced row echelon form (also called row canonical form) if it satisfies the following conditions:

  • It is in row echelon form.
  • Every leading coefficient is 1 and is the only nonzero entry in its column.

I suspect that is what you are asking. The main difference is that one can reach row-echelon form more quickly. The canonical form of row-echelon form is reduced row-echelon may require additional elementary row operations.

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