Given a system, we first check the consistency. If the system is not consistent, then we cannot say anymore about the solutions. If the system is consistent, then we may further ask specific questions about the solutions.

In the subsequent discussion, we always assume the systems are consistent.

Example Suppose the augmented
matrix of a system ** Ax** =

[ | 5 | 2 | 1 | -9 | 0 | -3 | ] |

0 | 0 | -7 | 1 | 4 | 1 | ||

0 | 0 | 0 | 3 | 5 | -2 |

The last equation is 3`x`_{4} + 5`x`_{5}
= -2. Because the pivot entry 3 (the coefficient of `x`_{4}) is
nonzero, we can move terms other than `x`_{4} to the right side
and then divide 3. The result is an expression of `x`_{4} in terms
of `x`_{5}: `x`_{4} = - 2/3 - 5/3 `x`_{5}.

The middle equation is -7`x`_{3} + `x`_{4}
+ 4`x`_{5} = 1. Because the pivot entry -7 (the coefficient of
`x`_{3}) is nonzero, we can move terms other than `x`_{3}
to the right side and then divide -7. The result is an expression of `x`_{3}
in terms of `x`_{4} and `x`_{5}. By further substituting
the earlier expression `x`_{4} = - 2/3 - 5/3 `x`_{5},
we see that `x`_{3} is expressed in terms of `x`_{5}.

Similarly, in the top equation, because 5 (the coefficient
of `x`_{1}) is a pivot, we can move terms other than `x`_{1}
to the right side and then divide the coefficient 5. The result is an expression
of `x`_{1} in terms of `x`_{2}, `x`_{3},
`x`_{4}, `x`_{5}. By further substituting the earlier
expressions of `x`_{3} and `x`_{4} in terms of `x`_{5},
we see that `x`_{1} is expressed in terms of `x`_{2}
and `x`_{5}.

The key observation here is that the pivot coefficient can be divided because it is *nonzero*. As a result, the corresponding
"pivot variable" can be expressed in terms of "later variables".
The back substitution process then further replace the "later variables"
with "nonpivot variables".

Example In this example, we studied a system with the augmented matrix having the following row echelon form.

[ | * | # | # | # | # | ] |

0 | 0 | * | # | # | ||

0 | 0 | 0 | * | # | ||

0 | 0 | 0 | 0 | 0 |

By back substitution and successively dividing the coefficients
of `x`_{4}, `x`_{3}, `x`_{1}, we may
write the general solution as expressions of `x`_{1}, `x`_{3},
`x`_{4} in terms of `x`_{2}, with `x`_{2}
arbitrary. Alternatively, by further row operations, we have the reduced
row echelon form.

[ | 1 | a |
0 | 0 | b_{1} |
] |

0 | 0 | 1 | 0 | b_{2} |
||

0 | 0 | 0 | 1 | b_{3} |
||

0 | 0 | 0 | 0 | 0 |

Then the general solution of the system is `x`_{1}
= `b`_{1} - `ax`_{2}, `x`_{3} = `b`_{2},
`x`_{4} = `b`_{3}, with `x`_{2} arbitrary.

Further examples can be found here. It is easy to conclude from the examples that, for a consistent system, the general solution can be presented as expressions of some nonfree variables (which cannot be arbitrarily chosen) in terms of some free variables (which can be arbitrarily chosen). Moreover, under the correspondence

variables of system ⇔ columns of coefficient matrix,

we have

Example In an earlier example, we considered a system

x_{1} |
+ 3x_{2} |
+ 2x_{3} |
+ x_{5} |
= | 0 | |

- x_{1} |
- x_{2} |
- x_{3} |
+ x_{4} |
= | 1 | |

4x_{2} |
+ 2x_{3} |
+ 4x_{4} |
+ 3x_{5} |
= | 2 | |

x_{1} |
+ 3x_{2} |
+ 2x_{3} |
- 2x_{4} |
= | 0 |

The row echelon form of the augmented matrix is

[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |

0 | 2 | 1 | -1 | 0 | 1 | ||

0 | 0 | 0 | 2 | 1 | 0 | ||

0 | 0 | 0 | 0 | 0 | 0 |

which implies the system is consistent. Moreover, the solution
is not unique because we may express nonfree variables `x`_{1},
`x`_{2}, `x`_{4} in terms of free variables `x`_{3},
`x`_{5}. Later in the same example, we took away the freedom `x`_{3},
`x`_{5} (i.e., assuming both are zero) and considered the system

x_{1} |
+ 3x_{2} |
= | 0 | |

- x_{1} |
- x_{2} |
+ x_{4} |
= | 1 |

4x_{2} |
+ 4x_{4} |
= | 2 | |

x_{1} |
+ 3x_{2} |
- 2x_{4} |
= | 0 |

The row echelon form of the augmented matrix is

[ | 1 | 3 | 0 | 0 | ] |

0 | 2 | -1 | 1 | ||

0 | 0 | 2 | 0 | ||

0 | 0 | 0 | 0 |

The system is still consistent. Since all the columns of the coefficient matrix (i.e., the first three columns) are pivot, all the variables are nonfree, so that the solution is unique. The uniqueness is to be expected because all the freedoms have been taken away.