### Proof of Properties

##### 3. Justification of geometry

In this section, several properties of the determinant are given. Moreover, all except the transpose property were given geometrical meaning. The transpose property has been proved here. Below we will prove the other properties.

Product Property Let `A` and `B` be square matrices of the same size, then `det`**AB** = `det`**A** det**B**.

We fix `A` and consider` D`(`B`) = `det`**AB** as a function of `B`. The following shows that `D` preserves addition in columns of `B`.

`D`( ..., **u** + `v`, ... )

= `det`(`A`[..., **u** + `v`, ...]) (definition of `D`)

= `det`[..., `A`(**u** + `v`), ...] (this formula)

= `det`[..., **Au** + `Av`, ...]

= `det`[..., **Au**, ...] + `det`[..., `Av`, ...] (`det` is multilinear)

= `det`(`A`[..., **u**, ...]) + `det`(`A`[..., **u**, ...]) (this formula)

= `D`( ..., **u**, ... ) + `D`( ..., `v`, ... ).
(definition of `D`)

Similarly, we may verify that `D` preserves scalar multiplication in columns of `B`. Moreover, the following shows `D` is alternating in columns of `B`.

`D`( ..., **v**, ..., `u`, ... )

= `det`(`A`[..., **v**, ..., `u`, ...]) (definition of `D`)

= `det`[..., **Av**, ..., `Au`, ...] (this formula)

= - `det`[..., **Au**, ..., `Av`, ...] (`det` is alternating)

= -`det`(`A`[..., **u**, ..., `v`, ...]) (this formula)

= `D`( ..., **u**, ..., `v`, ... ). (definition of `D`)

Thus by this result, we have

`det`**AB** = `D`(`B`) = `D`(`I`) `det`**B** = `det`**AI** det**B** = `det`**A** det**B**.

Triangle Property Let `A` and `B` be square matrices. Then `det`[ |
`A` |
`O` |
] = `det`**A** det**B**. |

**#** |
**B** |

We may use the idea similar to the proof of the product property. For fixed `A`, it is easy to verify that the function

is multilinear and alternating in columns of `B`. Then by this result, we have `D`(`B`) = `D`(`I`) `det`**B**. Now

may be computed by repeated use of cofactor expansion (already proved here) along the last column. The result is `D`(`I`) = `det`**A**. This completes the proof that `D`(`B`) = `det`**A** det**B**.