Exercise Find invertible matrices and compute the inverse.

1)

[ | 3 | -2 | ] |

5 | -3 |

Answer By the conclusion about the inverse of 2 by 2 matrices, since 3×(-3) - 5×(-2) ≠ 0, the matrix is invertible. The inverse is

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

-5 | 3 | -5 | 3 |

2)

[ | 3 | 5 | ] |

-2 | -3 |

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

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

2 | 3 | 2 | 3 |

3)

[ | 6 | 4 | ] |

9 | 6 |

Answer Since 6×6 - 4×9 = 0, the matrix is not invertible.

4)

[ | 6 | 4 | ] |

9 | 7 |

Answer Since 6×7 - 4×9 ≠ 0, the matrix is invertible. The inverse is

(6×7 - 4×9)^{-1}[ |
7 | -4 | ] = [ | 7/6 | -2/3 | ] |

-9 | 6 | -3/2 | 1 |

5 - 8)

[ | 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 |

Answer The first three matrices are obtained by selecting columns from the fourth matrix. In an earlier exercise, we have computed the row echelon form

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

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

0 | 0 | 0 | 3 | -1 |

for the fourth matrix. The first matrix, being [col 1], [col 2], [col 4], can be row changed to

[ | 1 | 1 | 0 | ] |

0 | 1 | 1 | ||

0 | 0 | 3 |

which shows that the first matrix is invertible. To actually compute the inverse, we use this method. So we append the first matrix by the identity matrix.

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

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

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

The same row operations as in the earlier exercise give us

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

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

0 | 0 | 3 | 1 | -2 | 2 |

Then the operations 1/3[row 3], - [row 3] + [row 2], - [row 2] + [row 1] gives us

[ | 1 | 0 | 0 | 1/3 | 1/3 | -1/3 | ] |

0 | 1 | 0 | -1/3 | 2/3 | 1/3 | ||

0 | 0 | 1 | 1/3 | -2/3 | 2/3 |

From this we conclude

[ | 2 | 0 | 1 | ]^{-1} = [ |
1/3 | 1/3 | -1/3 | ] |

1 | 1 | 0 | -1/3 | 2/3 | 1/3 | |||

0 | 1 | 1 | 1/3 | -2/3 | 2/3 |

Similarly, the second matrix, being [col 1], [col 3], [col 4], has the following row echelon form

[ | 1 | -1 | 0 | ] |

0 | -2 | 1 | ||

0 | 0 | 3 |

which shows that the first matrix is invertible. To compute the inverse, we again append the identity matrix.

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

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

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

The same row operations give us

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

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

0 | 0 | 3 | 1 | -2 | 2 |

Then the operations 1/3[row 3], - [row 3] + [row 2], (-1/2)[row 2], [row 2] + [row 1] gives us

[ | 1 | 0 | 0 | 1/6 | 2/3 | -1/6 | ] |

0 | 1 | 0 | 1/6 | -1/3 | -1/6 | ||

0 | 0 | 1 | 1/3 | -2/3 | 2/3 |

From this we conclude

[ | 2 | 0 | 1 | ]^{-1} = [ |
1/6 | 2/3 | -1/6 | ] |

1 | 1 | 0 | 1/6 | -1/3 | -1/6 | |||

0 | 1 | 1 | 1/3 | -2/3 | 2/3 |

The third matrix has the following row echelon form

[ | 1 | 1 | -1 | ] |

0 | 1 | -2 | ||

0 | 0 | 0 |

which shows that the matrix is not invertible. Finally, the fourth matrix is not invertible because it is not a square matrix.

9)

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

2 | 1 | 1 | 0 | ||

0 | 3 | 1 | 0 | ||

3 | 0 | 1 | 0 |

Answer In an earlier exercise, we computed the row echelon form.

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

0 | 3 | 1 | 0 | ||

0 | 0 | 0 | 1 | ||

0 | 0 | 0 | 0 |

Since [col 3] (and [row 4]) are not pivot, by the criterion for the invertibility, the matrix is not invertible.

10)

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

0 | 1 | 2 | 3 | ||

0 | 0 | 1 | 2 | ||

0 | 0 | 0 | 1 |

Answer Since this is a square matrix with all rows (and columns) pivot, the matrix is invertible. To compute the inverse, we form

[ | 1 | 2 | 3 | 4 | 1 | 0 | 0 | 0 | ] |

0 | 1 | 2 | 3 | 0 | 1 | 0 | 0 | ||

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

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

Then -2[row 4] + [row 3], -3[row 4] + [row 2], -4[row 4] + [row 1] give us

[ | 1 | 2 | 3 | 0 | 1 | 0 | 0 | -4 | ] |

0 | 1 | 2 | 0 | 0 | 1 | 0 | -3 | ||

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

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

The operations -2[row 3] + [row 2], -3[row 3] + [row 1] further give us

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

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

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

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

The operations -2[row 2] + [row 1] finally give us

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

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

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

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

We conclude

[ | 1 | 2 | 3 | 4 | ]^{-1} = [ |
1 | -2 | 4 | 0 | ] |

0 | 1 | 2 | 3 | 0 | 1 | -2 | 1 | |||

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

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

11)

[ | 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 |

Answer Since the matrix is not square, it is not invertible.

12)

[ | 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 |

Answer We append the identity matrix

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

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

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

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

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

By -[row 1] + [row 2], -[row 1] + [row 3], -[row 1] + [row 4], -[row 1] + [row 5], we get

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

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

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

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

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

By adding [row 2] through [row 5] to [row 1], we get

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

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

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

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

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

By multiplying -1 to [row 2] through [row 5], we get

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

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

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

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

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

Thus we conclude

[ | 1 | 1 | 1 | 1 | 1 | ]^{-1} = [ |
-3 | 1 | 1 | 1 | 1 | ] |

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

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

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

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

Remark One can see the general pattern and conclude

[ | 1 | 1 | 1 | . . | 1 | ]^{-1} = [ |
2 - n |
1 | 1 | . . | 1 | ] |

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

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

: | : | : | : | : | : | : | : | |||||

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