### Geometry of Determinant

##### 1. Determinant of 2 by 2 matrix

The determinant of a 1 by 1 matrix is `det`(`a`) = `a`.

The determinant of a 2 by 2 matrix is

`det`[ |
`a` |
`b` |
] = `ad` - `bc`. |

`c` |
`d` |

The expression appeared in the criterion for the invertibility of a 2 by 2 matrix: A 2 by 2 matrix `A` is invertible ⇔ the determinant is nonzero. The criterion clearly also holds for a 1 by 1 matrix, and will be shown to be true for general `n` by `n` matrices.

Let `u` = (`a`, `c`), `v` = (`b`, `d`) be the two columns of the matrix. Then the determinant `det`[`u v`] is a function of two vectors.

The absolute value of `det`[`u v`] is the area of the parallelogram spanned by `u` and `v`.

The claim can be seen through the following picture.

The parellelogram is obtained by substracting four triangles from a rectangle. Thus

area of the parallelogram = (`a` + `b`)(`c` + `d`) - (`a` + `b`)`c` - `b`(`c` + `d`) = `ad` - `bc`.

Note that the argument is done for the vectors in the given positions, and the determinant turns out to be positive for the given situation. One may also carry out similar arguments for the other situations. See this exercise.

Example The area of the triangle with the vertices (1, 2), (7, 1), (3, 5) is half of the area of the parallelogram spanned by `u` = (7, 1) - (1, 2) = (6, -1) and `v` = (3, 5) - (1, 2) = (2, 3). Thus

area = (1/2)| `det`[ |
6 |
2 |
] | = 10. |

-1 |
3 |

The area of the 4-gon with the vertices (1, 2), (5, 1), (7, 6), (3, 5) is the sum of two triangles. Thus

area = (1/2)| `det`[ |
4 |
6 |
] | + (1/2)| `det`[ |
6 |
2 |
] | = 17. |

-1 |
5 |
5 |
3 |

A real number is determined by the absolute value and the sign. Thus the determinant of a 2 by 2 matrix would be completely understood if we also know the meaning of the sign.

`u` moves to `v` in counterclockwise direction ⇒ `det`[`u v`] ≥ 0.

`u` moves to `v` in clockwise direction ⇒ `det`[`u v`] ≤ 0.

In other words, the sign is given by the orientation from the first colunm vector to the second.

To see why the orientation determines the sign, we consider the following movie for the counterclockwise situation.

Since the vectors are changed continuously, the determinant is also changed continuously. Moreover, the determinant never becomes zero since the absolute value, being the area of the parallelogram, is obviously never zero. Since a nonzero continuous function never changes sign, we conclude that `det`[`u v`] and `det`[`e`_{1} `e`_{2}] = 1 have the same sign. In other words, `det`[`u v`] > 0. For the clockwise situation, one may construct a similar movie and conclude that `det`[`u v`] and `det`[`e`_{1} -`e`_{2}] = -1 have the same sign.

What about the case `det`[`u v`] = 0? In this case the area of the parallelogram is zero. This can happen only if `u` and `v` are *parallel*. By an earlier discussion, this is equivalent to the vectors are linearly dependent. Then by this result, this is also equivalent to that the matrix [`u v`] is not invertible. Thus we have a geometrical proof of the earlier criterion for the invertibility of a 2 by 2 matrix.