Remark Please do the problems by yourself and then compare with the given solution. Pay attention to the order the simplification is done: one column at a time, from left to right.

Please also note the three possible outcomes for the solutions: (1) no solution, (2) unique solution, (3) non-unique solution (with some arbitrary choice).

Exercise Solve systems of linear equations.

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

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

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

Answer The augmented matrix is

[ | 1 | -1 | 3 | 3 | ] |

3 | 1 | 1 | 5 | ||

1 | 1 | -1 | 1 |

We start the simplification with the [col 1]. By (-1)[row 3] + [row 1] and (-3)[row 3] + [row 2], [col 1] is as simple as we can get.

[ | 0 | -2 | 4 | 2 | ] |

0 | -2 | 4 | 2 | ||

1 | 1 | -1 | 1 |

Then the exchange [row 1] ↔ [row 3] puts the most complicated row at the top.

[ | 1 | 1 | -1 | 1 | ] |

0 | -2 | 4 | 2 | ||

0 | -2 | 4 | 2 |

Now we turn our attention to [col 2]. By (-1)[row 2] + [row 3], we create one more zero in [col 2].

[ | 1 | 1 | -1 | 1 | ] |

0 | -2 | 4 | 2 | ||

0 | 0 | 0 | 0 |

Incidentally, [col 3] and [col 4] are also simplified. In fact, the shape is already the simplest we can get. Converted to a system of linear equations, we have

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

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

0 | = | 0 |

Note that the last equation is trivial. The simplest nontrivial equation is the 2nd one, from which we get

`x`_{2} = -1 + 2`x`_{3},
where `x`_{3} is arbitrary.

Substituting this into the 1st equation, we get

`x`_{1} = 1 - `x`_{2}
+ `x`_{3} = 3 - 1 + 2`x`_{3} - 3`x`_{3}
= 2 - `x`_{3}.

Thus the general solution is given by the formulae that express
`x`_{1} and `x`_{2} in terms of `x`_{3},
with `x`_{3} arbitrary.

2)

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

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

Answer The augmented matrix is

[ | 2 | -3 | -1 | 3 | ] |

1 | 1 | 2 | -1 |

We apply (-2)[row 2] + [row 1] to get

[ | 0 | -5 | -5 | 5 | ] |

1 | 1 | 2 | -1 |

Then by [row 1] ↔ [row 2] and (-1/5)[row 2], we have

[ | 1 | 1 | 2 | -1 | ] |

0 | 1 | 1 | -1 |

[row 2] corresponds to `x`_{2} + `x`_{3} = -1,
which gives us `x`_{2} = -1 - `x`_{3}, where `x`_{3} is arbitrary.
[row 1] corresponds to `x`_{1} + `x`_{2} + 2`x`_{3} = -1.
The general solution is

`x`_{1} = - 1 - `x`_{2} - 2`x`_{3}
= - 1 - (-1 - `x`_{3}) - 2`x`_{3} = - `x`_{3}

`x`_{2} = -1 - `x`_{3}

`x`_{3} is arbitrary

3)

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

4x_{1} |
+ 5x_{2} |
= | 6 |

7x_{1} |
+ 8x_{2} |
= | 9 |

Answer The augmented matrix is

[ | 1 | 2 | 3 | ] |

4 | 5 | 6 | ||

7 | 8 | 9 |

By (-4)[row 1] + [row 2] and (-7)[row 1] + [row 3], we get

[ | 1 | 2 | 3 | ] |

0 | -3 | -6 | ||

0 | -6 | -12 |

Then we apply (-2)[row 2] + [row 3] and (-1/3)[row 2] to get

[ | 1 | 2 | 3 | ] |

0 | 1 | 2 | ||

0 | 0 | 0 |

The last row corresponds to the trivial equality 0 = 0. The simplest nontrivial
row is the second one, corresponding to the equation `x`_{2} =
2. Substituting to the equation `x`_{1} + 2`x`_{2}
= 3, we have `x`_{1} = 3 - 4 = 1. Thus the system has a unique solution
`x`_{1} = 1, `x`_{2} = 2.

4&5)

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

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

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

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

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 |

Answer
The two systems have the same coefficient matrix ` A`, with different right sides

[] = [A b c |
1 | 3 | 2 | 0 | 1 | 0 | 0 | ] |

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

0 | 4 | 2 | 4 | 3 | 3 | 2 | ||

1 | 3 | 2 | -2 | 0 | 0 | 0 |

By first applying [row 2] + [row 4] and then applying [row 1] + [row 2], we have

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

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

0 | 4 | 2 | 4 | 3 | 3 | 2 | ||

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

Next we apply (-2)[row 2] + [row 3] and (-1)[row 2] + [row 4] to get

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

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

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

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

Finally we apply [row 3] + [row 4] to get

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

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

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

0 | 0 | 0 | 0 | 0 | 1 | 0 |

If we take away the last column, then we see that the row operation changes the augmented matrix of the 4th system

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

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

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

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

into

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

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

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

0 | 0 | 0 | 0 | 0 | 1 |

The last row corresponds to the equation 0 = 1. The contradiction shows that the fourth system has no solution.

Similarly, by taking away the 6th column, we see that the row operation changes the augmented matrix of the 5th system

x_{1} |
+3 x_{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 |

into

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

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

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

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

The last equation is trivial. The 3rd equation gives us

`x`_{4} = 1/2 - 1/2`x`_{5}, `x`_{5} arbitrary.

The 2nd equation gives us

`x`_{2} = 1/2 - 1/2`x`_{3} - 1/2`x`_{4} - 1/2`x`_{5}
= 1/4 - 1/2`x`_{3} - 1/4`x`_{5}, `x`_{3} arbitrary.

The 1st equation then gives us

`x`_{1} = - 3`x`_{2} - 2`x`_{3} - `x`_{5}
= - 3/4 - 1/4`x`_{3} - 1/4`x`_{5}.

The general solution is given by the expressions of `x`_{1}, `x`_{2},
`x`_{4} in terms of arbitrary `x`_{3}, `x`_{5}.