Exercise For what choices of the parameters, the following matrices are invertible?

1)

[ | a + 1 |
a - 1 |
] |

a |
a + 2 |

Answer The condition for the matrix to be invertible is
(`a` + 1)(`a` + 2) - `a`(`a` - 1) ≠ 0. Since
(`a` + 1)(`a` + 2) - `a`(`a` - 1) = 4`a` + 2, we see that the condition is `a` ≠ -1/2.

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

1 | 0 | 0 | 0 | ||

2 | 2 | 0 | a |
||

0 | 1 | 3 | a |

Answer The matrix appears as [col 1], [col 2], [col 4], [col 5] of the matrix in an earlier exercise. According to the computation there, row operations change the matrix into

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

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

0 | 0 | -2 | a - 2 |
||

0 | 0 | 0 | 2a - 3 |

By the criterion, we need all columns (and rows) to be pivot in order for
the matrix to be invertible. Thus the condition is `a` ≠ 3/2.

3)

[ | 1 | 1 | -1 | ] |

1 | 0 | -a |
||

1 | a |
0 |

Answer The matrix appears as the coefficient matrix of the system in an earlier exercise. According to the computation there, row operations change the matrix into

[ | 1 | 1 | -1 | ] |

0 | -1 | 1 - a |
||

0 | 0 | 2a - a^{2} |

The condition for the matrix to be invertible is that all columns (and rows) are pivot,
which means 2`a` - `a`^{2} ≠ 0, i.e., `a` ≠ 0 or 2.

[ | 0 | 1 | 1 | a |
] |

1 | 0 | a |
1 | ||

1 | 1 | 0 | 0 | ||

0 | a |
-1 | 1 |

Answer According to the computation in an earlier exercise, row operations change the matrix into

[ | 1 | 0 | a |
1 | ] |

0 | 1 | 1 | a |
||

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

0 | 0 | 0 | -a^{2} + a + 2 |

The condition for the matrix to be invertible is - `a` - 1 ≠ 0 and
- `a`^{2} + `a` + 2 ≠ 0, which means `a` ≠ -1 or 2.

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

a |
1 | 0 | 0 | ||

0 | a |
1 | 0 | ||

0 | 0 | a |
1 |

Answer By (-`a`)[row 1] + [row 2], (-`a`)[row 2] + [row 3],
(-`a`)[row 3] + [row 4], the matrix becomes the identity. Therefore the matrix is invertible for any `a`.

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

a |
1 | 1 | 1 | ||

a |
a |
1 | 1 | ||

a |
a |
a |
1 |

Answer By (-`a`)[row 1] + [row 2], (-`a`)[row 1] + [row 3],
(-`a`)[row 1] + [row 4], the matrix becomes

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

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

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

0 | 0 | 0 | 1 - a |

Therefore the matrix is invertible for `a` ≠ 1.

[ | a |
b |
b |
b |
] |

a |
a |
b |
b |
||

a |
a |
a |
b |
||

a |
a |
a |
a |

Answer By - [row 3] + [row 4], - [row 2] + [row 3], - [row 1] + [row 2], the matrix becomes

[ | a |
b |
b |
b |
] |

0 | a - b |
0 | 0 | ||

0 | 0 | a - b |
0 | ||

0 | 0 | 0 | a - b |

Therefore the matrix is invertible for `a` ≠ 0, `b`.