Elimination can go further than a staircase. Scale every pivot to 1 and clear the entries above each pivot as well as below it, and the solutions can be read straight off the matrix, with no back substitution.

On a tidy desk each thing has its own place and nothing is stacked on it. In the reduced form each pivot column is clean: a single 1 and zeros everywhere else.

A matrix is in reduced row echelon form when it is in row echelon form, every pivot is $1$, and every pivot is the only non-zero entry of its column. Gauss–Jordan elimination reaches it: after the forward pass, scale each pivot row to make its pivot $1$, then clear the entries above the pivots, from the last pivot back to the first. Every matrix has exactly one reduced row echelon form, whatever operations are used to reach it; that uniqueness is stated here and used, not proved.

• Each non-zero row names one basic variable and expresses it through the free ones. Above, in $x_1,x_2,x_3,x_4$: $x_1=2-3x_3$, $x_2=5+x_3$ and $x_4=-4$, with $x_3$ free.
• Because the reduced form is unique, the pivot positions of a matrix, and so the number of its pivots, do not depend on how elimination was done.
• An $n\times n$ matrix whose reduced form is $I_n$ gives $A\mathbf{x}=\mathbf{b}$ exactly one solution for every $$: the one read off the last column.
• Gauss–Jordan costs more arithmetic than back substitution for a single system, but the reduced form is what makes inverses and many later computations easy to read.

• It is in row echelon form.
• Every pivot is $1$.
• Each pivot is the only non-zero entry in its column.

• Reduced row echelon form: echelon form, every pivot $1$, each pivot alone in its column.
• Gauss–Jordan elimination: the forward pass, then scale the pivots, then clear upwards.
• The reduced form is unique, so pivot positions are fixed.
• Each non-zero row gives one basic variable in terms of the free ones.

Great work! You now know how to:

• recognise and reach reduced row echelon form;
• read the solutions straight from it;
• use its uniqueness: pivot positions belong to the matrix;
• recognise $I_n$ as the reduced form of a square matrix with $n$ pivots.