Exercise Find invertible matrices and compute the inverse.

[ | 3 | -2 | ] |

5 | -3 |

[ | 3 | 5 | ] |

-2 | -3 |

[ | 6 | 4 | ] |

9 | 6 |

[ | 6 | 4 | ] |

9 | 7 |

[ | 2 | 0 | 1 | ] |

1 | 1 | 0 | ||

0 | 1 | 1 |

[ | 2 | 2 | 1 | ] |

1 | -1 | 0 | ||

0 | -2 | 1 |

[ | 2 | 0 | 2 | ] |

1 | 1 | -1 | ||

0 | 1 | -2 |

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

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

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

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

2 | 1 | 1 | 0 | ||

0 | 3 | 1 | 0 | ||

3 | 0 | 1 | 0 |

[ | 1 | 2 | 3 | 4 | ] |

0 | 1 | 2 | 3 | ||

0 | 0 | 1 | 2 | ||

0 | 0 | 0 | 1 |

[ | 10.2 | 0.3 | -5.4 | ] |

-7.9 | 8.2 | 12.4 | ||

0.8 | -0.9 | 1.6 | ||

33.5 | -44.9 | 38.4 | ||

-5.9 | -7.3 | 0.1 |

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

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

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

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

1 | 1 | 1 | 1 | 0 |

Exercise Find invertible linear transformations and compute the inverse.

**T**_{1}(`x`_{1}, `x`_{2})
= (2`x`_{1} + 3`x`_{2}, 5`x`_{1} + 7`x`_{2}):
**R**^{2} → **R**^{2}

**T**_{2}(`x`, `y`)
= (2`x` - 6`y`, - 5`x` + 15`y`):
**R**^{2} → **R**^{2}

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

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

**T**_{5}(`x`_{1}, `x`_{2}, `x`_{3})
= (- 2`x`_{2} + `x`_{3},
`x`_{1} - `x`_{3},
3`x`_{1} + 2`x`_{2} - 4`x`_{3}):
**R**^{3} → **R**^{3}

**T**_{6}(`x`_{1}, `x`_{2}, `x`_{3}, `x`_{4})
= (`x`_{2} + 2`x`_{3} + 4`x`_{3},
`x`_{1} + `x`_{2} + 2`x`_{3},
`x`_{1} + `x`_{2} + 2`x`_{3} - 2`x`_{4},
`x`_{1} + `x`_{2} + `x`_{3} + 2`x`_{4}):
**R**^{4} → **R**^{4}

**T**_{7}(`x`_{1}, `x`_{2}, `x`_{3}, `x`_{4}, `x`_{5})
= (`x`_{1} + `x`_{2},
`x`_{2} + `x`_{3},
`x`_{3} + `x`_{4},
`x`_{4} + `x`_{5},
`x`_{5} + `x`_{1}):
**R**^{5} → **R**^{5}

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

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

a |
a + 2 |

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

1 | 0 | 0 | 0 | ||

2 | 2 | 0 | a |
||

0 | 1 | 3 | a |

[ | 1 | 1 | -1 | ] |

1 | 0 | -a |
||

1 | a |
0 |

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

1 | 0 | a |
1 | ||

1 | 1 | 0 | 0 | ||

0 | a |
-1 | 1 |

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

a |
1 | 0 | 0 | ||

0 | a |
1 | 0 | ||

0 | 0 | a |
1 |

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

a |
1 | 1 | 1 | ||

a |
a |
1 | 1 | ||

a |
a |
a |
1 |

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

a |
a |
b |
b |
||

a |
a |
a |
b |
||

a |
a |
a |
a |

Exercise Compute the inverse of the following matrices.

[ | 2 | 3 | ] |

7 | 9 |

[ | 1 | 0 | 2 | ] |

2 | -1 | 3 | ||

4 | -1 | 8 |

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

1 | 1 | -1 | 0 | ||

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

1 | -1 | 2 | 1 |

Then use your result to solve the following systems.

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

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

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

7x_{1} |
+ 9x_{2} |
= | 0 |

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

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

4x_{1} |
- x_{2} |
+ 8x_{3} |
= | -1 |

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

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

4x_{1} |
- x_{2} |
+ 8x_{3} |
= | 1 |

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

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

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

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

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**)

Exercise What is the condition for a diagonal matrix to be invertible? What is the inverse? What about an upper (lower) triangular matrix?

Exercise Prove that

[ | a |
b |
] |

c |
d |

is invertible ⇔ `ad` - `bc` ≠ 0.

Exercise Prove that if row operations change [** A B**] to
[

Exercise Let ` A` and

Exercise

- Prove that
=`Ax`has solutions for any**b**⇔ There is a matrix**b**, such that**X**=`AX`.**I** - Prove that solution of
=`Ax`is unique ⇔ There is a matrix**b**, such that**X**=`XA`.**I** - Rephrase the first and second parts in terms of linear transformations.