Elimination aims at a staircase. Each row starts further right than the one above it, and below each step there are only zeros. A system in that shape is solved from the bottom up, one unknown at a time.

A staircase is climbed one step at a time from the bottom. Back substitution does the same: the last equation gives one unknown, and each step up uses what is already known.

A matrix is in row echelon form when every row of zeros lies at the bottom, and the first non-zero entry of each other row, its leading entry, lies strictly to the right of the leading entry of the row above. These leading entries are the pivots, and the columns holding them are the pivot columns. Forward elimination reaches this form; it is the method called Gaussian elimination.

• Work column by column from the left. In the current column, find a row at or below the current one with a non-zero entry, and exchange it upwards if needed. That entry is the pivot.
• Subtract multiples of the pivot row from every row below it, so that every entry under the pivot becomes zero.
• Move one row down and one column right, and repeat. If a column has no non-zero entry at or below the current row, it gets no pivot: move right without moving down.
• Back substitution: solve the last non-zero equation for its pivot unknown, substitute upwards, and continue to the top.
• A matrix has many echelon forms, depending on the choices made along the way, but the positions of its pivots are always the same. A later lesson explains why.

• All rows of zeros are at the bottom.
• Each leading entry lies strictly to the right of the one above it.
• So every entry below a leading entry is zero.

The first pivot is the top left entry, if it is not zero. Clearing the column under it takes one row operation for each row below.

In echelon form the last equation has one unknown. Solve it, put the value into the equation above, solve that, and climb.

• Row echelon form: zero rows at the bottom, each leading entry right of the one above.
• The leading entries are the pivots; their columns are the pivot columns.
• Gaussian elimination clears each column under its pivot, exchanging rows when the pivot position holds a zero.
• Back substitution solves from the bottom row up.

Great work! You now know how to:

• recognise row echelon form and its pivots;
• carry out Gaussian elimination, with exchanges when needed;
• solve by back substitution;
• know which features of an echelon form are fixed and which are choices.