### Inverse

Exercise Show that

**A** = [ |
1 |
0 |
-2 |
], **B** = [ |
-3 |
-2 |
2 |
] |

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

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

are inverse to each other. Then compute

**C** = **A**[ |
1 |
0 |
0 |
]**B** |

0 |
-1 |
0 |

0 |
0 |
0 |

and **C**^{2001}, (`AC`)^{-1}, **A**^{-2}.

Answer By this criterion,
to verify **A** and **B** are inverse to each other, it suffices to show `AB` = **I**.
A straightforward computation confirms this. Another straightforward computation gives us

**C** = [ |
-3 |
-2 |
2 |
] |

0 |
-1 |
0 |

-6 |
-5 |
4 |

Then by **C** = `ADB`, **C**^{2001} =
(`ADB`)(`ADB`)...(`ADB`) =
`AD`(`BA`)**D**(`BA`)**D**...**D**(`BA`)`DB` =
`ADIDID`...`DIDB` = `AD`^{2001}**B**,
and the obvious equality **D**^{2001} = **D**,
we conclude that **C**^{2001} = **C**.

By this equality, we have

(`AC`)^{-1} =
**C**^{-1}**A**^{-1} =
`CA`^{-1} = [ |
1 |
0 |
0 |
] |

0 |
-1 |
0 |

2 |
-1 |
0 |

where we used **C**^{-1} = (`ADB`)^{-1} =
**B**^{-1}**D**^{-1}**A**^{-1} =
`ADB` = **C**.

Finally, we have

**A**^{-2} =
**B**^{2} = [ |
5 |
2 |
-4 |
] |

0 |
1 |
0 |

1 |
2 |
-3 |