Recall that an invertible matrix ` A` must be a square matrix
such that

The justification of the method can be found later in the proof of criterion for invertibility.

Example To find the inverse of

= [A |
3 | 1 | -1 | ] |

1 | -1 | 1 | ||

2 | 2 | 1 |

we form

[] = [A I |
3 | 1 | -1 | 1 | 0 | 0 | ] |

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

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

The row operations corresponding to what we have done to ` A` in this
example and this
example give us

[ | 1 | -1 | 1 | 0 | 1 | 0 | ] |

0 | 4 | -4 | 1 | -3 | 0 | ||

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

Further row operations give us

[ | 1 | 0 | 0 | 1/4 | 1/4 | 0 | ] |

0 | 1 | 0 | -1/12 | -5/12 | 1/3 | ||

0 | 0 | 1 | -1/3 | 1/3 | 1/3 |

Thus we conclude

A^{-1} = [ |
1/4 | 1/4 | 0 | ] |

-1/12 | -5/12 | 1/3 | ||

-1/3 | 1/3 | 1/3 |

By the way, the vector

[ | 1/4 | 1/4 | 0 | ] [ | 2 | ] = [ | 1 | ] |

-1/12 | -5/12 | 1/3 | 2 | 1 | ||||

-1/3 | 1/3 | 1/3 | 6 | 2 |

is exactly the solution of the system in this example.

Exercise To find the inverse of

= [A |
1 | 0 | 1 | 0 | ] |

1 | 1 | 0 | 1 | ||

0 | 0 | 1 | 1 | ||

1 | 1 | 1 | 1 |

we form

[] = [A I |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | ] |

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

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

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

Row operations on the matrix give us

[ | 1 | 0 | 0 | 0 | 1 | 1 | 0 | -1 | ] |

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

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

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

Thus we have

A^{-1} = [ |
1 | 0 | 0 | 0 | 1 | 1 | 0 | -1 | ] |

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

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

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

In terms of linear transformations, the inverse of

` T`(

is

**T**^{-1}(`x`_{1}, `x`_{2}, `x`_{3}, `x`_{4})
= (`x`_{1} + `x`_{2} - `x`_{4},
- `x`_{1} - `x`_{2} - `x`_{3} + 2`x`_{4},
- `x`_{2} + `x`_{4},
`x`_{2} + `x`_{3} - `x`_{4}):
**R**^{4} → **R**^{4}