In the previous video, we saw how a system of simultaneous linear equations can be encoded as an augmented matrix and how manipulations of the equations corresponded to performing row operations on the augmented matrix. In the example we worked through, we started with: $$x-y=-1,x+y=3,\left(\begin{array}{cccc}\hfill 1\hfill & \hfill -1\hfill & \hfill |\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill |\hfill & \hfill 3\hfill \end{array}\right)$$
and ended up with $$x=1,y=2,\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill |\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill |\hfill & \hfill 2\hfill \end{array}\right).$$
Because the block on the left of the vertical bar in the augmented matrix is the identity, this tells us that the variables $x$
and $y$
are completely determined. It's not always possible to make a matrix equal to the identity by performing row operations. It is, however, always possible to put it into a special form called reduced echelon form. We'll see how this works in the next three videos.

Echelon form

Definition:

Suppose we have an $m$
-by-$n$
(augmented) matrix like: $$\left(\begin{array}{cccccc}\hfill 0\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill |\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill |\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill |\hfill & \hfill 4\hfill \end{array}\right)$$
For each nonzero row, the leading entry is the leftmost nonzero entry (left of the bar). In some row, if there are no nonzero entries left of the bar, then there is no leading entry in this row. In the example above, the leading entries are:

for the first row: $5$
,

for the second row: there is no leading entry (nothing nonzero to the left of the bar),

for the third row: $7$
.

The $n$
-by-$n$
identity matrix has leading entries $1,1,1,\mathrm{\dots},1$
. These move to the right as you work down the rows.

Definition:

A matrix is in echelon form if all the zero-rows are at the bottom and the leading entries move strictly to the right as you go down the rows.

Our earlier example is not in echelon form: there's a row of zeros in the middle, which should be at the bottom. Even if that weren't a problem, its leading entries ($5$
and $7$
) move to the left as you go down the rows.

The 4-by-3 matrix $\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$
is in echelon form. All zero-rows are at the bottom. The leading entries are ($1$
in the top row and $2$
in the second) move to the right, so this is in echelon form. The "staircase of zeros" in the bottom left of the matrix is the reason that echelon form is called echelon form: the French word échelle means ladder.

The 3-by-3 matrix $\left(\begin{array}{ccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$
is not in echelon form: the leading indices don't move strictly to the right (3 and 4 are in the same column instead of moving to the right.

Spot the echelons

Which of the following matrices are in echelon form?