The number of pivots is the single number that says most about a matrix, and it is called the rank. Compared with the number of unknowns and with the rank of the augmented matrix, it decides whether a system has no solution, exactly one, or infinitely many.

A committee of ten may hold only three different views if the other seven only repeat combinations of those three. The rank counts the rows that elimination cannot wipe out.

The rank of a matrix $A$, written $\operatorname{rank}A$, is the number of pivots in a row echelon form of $A$. Because the reduced form is unique, it does not depend on how elimination is done. For an $m\times n$ matrix it is at most $m$, since each pivot has its own row, and at most $n$, since each has its own column.

• $A\mathbf{x}=\mathbf{b}$ has a solution exactly when $\operatorname{rank}A=\operatorname{rank}[A\mid\mathbf{b}]$: the theorem below.
• If it has one, the number of free variables is $n-\mathrm{rank}$, where is the number of unknowns.

Reduce the augmented matrix. The part before the bar is then an echelon form of A, so the two ranks count the pivots of one elimination and differ only if the last column holds a pivot. Such a pivot is a row that reads 0 = c with c not zero, which nothing satisfies. Without it, back substitution produces a solution.

Reduce the augmented matrix to echelon form; the part before the bar is an echelon form $U$ of $A$.

The pivots before the bar are those of $A$, and the last column can add at most one more.

A pivot in the last column is a row that is zero before the bar and not after it.

That row is an equation no values of the unknowns can satisfy.

Otherwise every equation can be solved for its pivot unknown, with the free unknowns set to anything, say zero.

Reduce to echelon form and count the non-zero rows: each of them holds exactly one pivot.

Two non-zero rows, so the rank is $2$: the second row was twice the first and added nothing.

Compare $\operatorname{rank}A$ with $\operatorname{rank}[A\mid\mathbf{b}]$. Equal: solvable. Different: the augmented matrix has a pivot after the bar, a row $0=c$.

• $\operatorname{rank}A$ is the number of pivots, and it is at most the number of rows and of columns.
• Solvable exactly when $\operatorname{rank}A=\operatorname{rank}[A\mid\mathbf{b}]$.
• Consistent systems have $n-\operatorname{rank}A$ free variables.

Great work! You now know how to:

• compute the rank by counting pivots;
• decide consistency with the rank test;
• count free variables and solutions from the rank;
• tell when a homogeneous system has a non-zero solution.