Exercise Find invertible linear transformations and compute the inverse.

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

Answer The matrix for **T**_{1} is

[ | 2 | 3 | ] |

5 | 7 |

Since 2×7 - 3×5 ≠ 0, the matrix is invertible. The inverse is

(2×7 - 3×5)^{-1}[ |
7 | -3 | ] = [ | -7 | 3 | ] |

-5 | 2 | 5 | -2 |

Thus the inverse transformation is **T**_{1}^{-1}(`x`_{1}, `x`_{2})
= (- 7`x`_{1} + 3`x`_{2}, 5`x`_{1} - 2`x`_{2}).

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

Answer The matrix for **T**_{2} is

[ | 2 | -6 | ] |

-5 | 15 |

Since 2×15 - (-5)×(-6) = 0, the matrix is not invertible. Correspondingly, the linear transformation is not invertible.

3) **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}

Answer The matrix of the transformation is

[ | 0 | -2 | 1 | ] |

1 | 0 | -1 | ||

3 | 1 | -3 |

We append the identity matrix.

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

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

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

Row operations change this to

[ | 1 | 0 | 0 | 1 | -5 | 2 | ] |

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

0 | 0 | 1 | 1 | -6 | 2 |

Thus we have

[ | 0 | -2 | 1 | ]^{-1} = [ |
1 | -5 | 2 | ] |

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

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

and the inverse transformation is **T**_{3}^{-1}(`x`_{1}, `x`_{2}, `x`_{3})
= (`x`_{1}- 5`x`_{2} + 2`x`_{3},
- 3`x`_{2} + `x`_{3},
`x`_{1} - 6`x`_{2} + 2`x`_{3}).

4) **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}

Answer The transformation is not invertible because the domain and codomain have different dimensions (see here).

5) **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}

Answer The matrix of the transformation is

[ | 0 | -2 | 1 | ] |

1 | 0 | -1 | ||

3 | 2 | -4 |

Row operations change this to

[ | 1 | 0 | -1 | ] |

0 | -2 | 1 | ||

0 | 0 | 0 |

Thus the matrix, as well as the linear transformation, is not invertible.

6) **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}

Answer The corresponding matrix is

[ | 0 | 1 | 2 | 4 | ] |

1 | 1 | 2 | 0 | ||

1 | 1 | 2 | -2 | ||

1 | 1 | 1 | 2 |

The matrix is invertible, with

[ | -1 | 3 | -2 | 0 | ] |

1 | -6 | 4 | 2 | ||

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

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

as the inverse matrix. Thus
**T**_{6}^{-1}(`x`_{1}, `x`_{2}, `x`_{3}, `x`_{4})
= (- `x`_{1} + 3`x`_{2} - 2`x`_{3},
`x`_{1} - 6`x`_{2} + 4`x`_{3} + 2`x`_{4},
2`x`_{2} - `x`_{3} - `x`_{4},
1/2 `x`_{2} + 1/2 `x`_{3}).

7) **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}

Answer The corresponding matrix is

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

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

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

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

1 | 0 | 0 | 0 | 1 |

The matrix is invertible, with

1/2[ | 1 | -1 | 1 | -1 | 1 | ] |

1 | 1 | -1 | 1 | -1 | ||

-1 | 1 | 1 | -1 | 1 | ||

1 | -1 | 1 | 1 | -1 | ||

-1 | 1 | -1 | 1 | 1 |

as the inverse matrix. Thus **T**_{7}^{-1}(`x`_{1}, `x`_{2}, `x`_{3}, `x`_{4}, `x`_{5})
= 1/2(`x`_{1} - `x`_{2} + `x`_{3} - `x`_{4} + `x`_{5},
`x`_{1} + `x`_{2} - `x`_{3} + `x`_{4} - `x`_{5},
- `x`_{1} + `x`_{2} + `x`_{3} - `x`_{4} + `x`_{5},
`x`_{1} - `x`_{2} + `x`_{3} + `x`_{4} - `x`_{5},
- `x`_{1} + `x`_{2} - `x`_{3} + `x`_{4} + `x`_{5}).

Remark
For ` T`(

**T**_{-1}(`x`_{1}, `x`_{2}, ..., `x _{n}`)
=

1/2(

If `n` is even, then for the nonzero vector
(1, -1, 1, -1, ..., 1, -1), we have ` T`(1, -1, 1, -1, ..., 1, -1) =