Let ` T`:

` T`(

is a linear equation. The following tells us the structure of solutions of linear equations.

The solutions of a linear equation ` T`(

`x`_{0}is one particular solution of(**T**) =`x`.`b`is any vector in**x**_{h}`kernel`.**T**

Proof Let `x`_{0} be one particular solution of
` T`(

` T`(

where the first equality is because ** T** is linear,
and the second equality is because

` T`(

Geometrically, the result means that the general solution of
` T`(

The result explains what we saw in the three examples at the beginning of this chapter.

Example The system

x_{1} |
+ x_{2} |
- 5x_{4} |
= | 3 | |

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

is linear because the left side is a linear transformation

` T`(

By taking `x`_{2} = `x`_{4} = 0, we find a special solution
`x`_{0} = (3, 0, 4, 0) of the system. Moreover, the kernel of ** T**,
which is the general solution of the homogeneous system

x_{1} |
+ x_{2} |
- 5x_{4} |
= | 0 | |

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

consists of vectors of the form

` x_{h}` =
(-

Therefore the general solution is

** x** =

We emphasize that the general solution is the sum of the fixed part `x`_{0} = (3, 0, 4, 0) and
the free part ` x_{h}` =

Example The differential equation

`f`'' - 4`f`' + 3`f` = 3`t` - 1

is linear because the left side is a linear transformation. The general solution is

`f` = (`t` + 1) + `c`_{1}`e ^{t}`
+

The fixed part `f`_{0} = `t` + 1 is a particular solution of the equation. The free part
`f _{h}` =

`f`'' - 4`f`' + 3`f` = 0.

We also note that, being all the linear combinations of `e ^{t}` and

Example The matrix equation

[ | 1 | -2 | 3 | ] = X, for 2 by 3 matrix IX |

-2 | 3 | -4 |

is linear because the left side is a linear transformation. The general solution is

= [X |
1 | -2 | ] + c_{1}[ |
1 | 0 | ] + c_{2}[ |
0 | 1 | ]. |

0 | 0 | 2 | 0 | 0 | 2 | ||||

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

The fixed part

X_{0} = [ |
1 | -2 | ] |

0 | 0 | ||

0 | 0 |

is a special solution, and the free part

= X_{h}c_{1}[ |
1 | 0 | ] + c_{2}[ |
0 | 1 | ] |

2 | 0 | 0 | 2 | |||

1 | 0 | 0 | 1 |

is all the solutions of the homogeneous equation

[ | 1 | -2 | 3 | ] = X.O |

-2 | 3 | -4 |

[previous topic] [part 1] [part 2] [part 3] [next topic]