How is the formula of determinant of $3\times 3$ or greater matrices derived?

Question

I have recently learned how to find the determinant of square matrices that are great than $2\times 2$. That part I got it without any trouble.

However, one thing that bothers me a lot is the formula (says definition in my textbook) where the determinant of $3\times 3$ matrix is $$a_{11}a_{22}a_{33} +a_{12}a_{23}a_{31} +a_{31}a_{12}a_{23} - (a_{13}a_{22}a_{31} +a_{21}a_{12}a_{33} +a_{11}a_{32}a_{23}).$$

Similar pattern applies to any other square matrices.

I cant understand where it comes from or how it is derived? It just feels so random to me

Also, when I was searching online, I have found that the determinant is actually the volume. Can anyone explain why that is the case as well?

Answer 1

There is a theorem in algebra that states that there is a unique function $$F:M_n(\mathbb K)\to\mathbb K$$ that is alternate multilinear such that $F(I_n)=1.$ Here $M_n(\mathbb K)$ is the space of $n\times n$ matrices with coefficient (= entries) in a field $\mathbb K.$ If you don't know what a field is, then just take $\mathbb K=\mathbb R$ or $\mathbb K =\mathbb C.$ Both of these are examples of fields. Alternate means that if you inverse two columns, then $F$ changes signs, and multilinear means it is linear in each of the columns.

It can be proven that this function $F$ is given by the formula \begin{equation} F(A)=\sum\limits_{\sigma\in S_n}\mathrm{sign}(\sigma)\prod\limits_{i=1}^n A_{\sigma(i),i} \end{equation} for some matrix $A.$ Here $S_n$ stands for the set of all permutations of $\{1,\ldots,n\}.$ Setting $\det=F$ gives you a rigorous definition of what the determinant actually is. Briefly, what I mean is that the determinant is the unique function that satisfies the above conditions, and it can be computed using the formula above. Proof and details can be found here: https://en.wikipedia.org/wiki/Leibniz_formula_for_determinants

As others already did in their answers, I won't show you that this formula gives you the identities for $2\times 2$ and $3\times 3$ matrices.

For the relation with area/volume, please see here: why determinant is volume of parallelepiped in any dimensions

Answer 2

A very nice illustration for this determinant formula for $n=3$ is the Rule of Sarrus (in German also "Zaunregel"). This yields: $$ \det(M) =\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12}. $$ For $n\ge 4$ there are no such beautiful rules, but there is the Laplace expansion, which is a recursive formula.

Answer 3

The (to my knowledge) original definition of the determinant for a $n \times n$-matrix $A = (a_{ij})$ is given by

$$\det A = \sum_{\pi \in S_n} sgn(\pi) \cdot a_{1\pi(1)} \cdot \ldots \cdot a_{n\pi(n)}$$ where $S_n = \{ \pi : \{1,\ldots, n\} \to \{1,\ldots, n\} ~ | ~ \pi \text{ bijective }\} $ and $sgn(\pi)$ the is the parity of the permutation $\pi$, i.e. either $1$ or $-1$.

That means whenever we want to compute a determinant we have to think of all the ways to permute the numbers $\{1,\ldots, n\}$. For example if $n = 2$ then the only permutations are $\pi_1$ and $\pi_2$ where $$\pi_1(1) = 1, \pi_1(2) = 2$$ $$\pi_2(1) = 2, \pi_2(2) = 1$$ Hence, the determinant of a $2 \times 2$-matrix is given by

$$\det \pmatrix{a_{11} & a_{12} \\ a_{21} & a_{22}} = sgn(\pi_1) a_{1\pi_1(1)}a_{2\pi_1(2)} + sgn(\pi_2) a_{1\pi_2(1)}a_{2\pi_2(2)} = a_{11}a_{22} - a_{12}a_{21}$$

Applying the formula to $n=3$ yields the formula you gave and can be illustrated quite nicely using diagonals of the matrix.

Answer 4

determinant of a $n\times n$ can be thought of as a $n$-linear alternating form over the set of all $n\times n$ matrices with the property $det(I)=1$ thus the conclusion follows. for a better understanding you can check "Linear algebra" by Hoffman & kunze

Answer 5

For an intuitive presentation you can check Gilbert Strang "Introduction to linear algebra".

The determinant of a square matrix is defined as a number which satisfies the following basic properties:

a. $det(I) = 1$

b. If you exchange 2 rows of a matrix, the absolute value of the determinant does not change, only the sign is reversed

c. The determinant is linear in each row:

• c1- if you multiply one row of a matrix by k, the determinant is multiplied by the same factor: $$ \begin{vmatrix} k*a & k*b \\ c & d \\ \end{vmatrix} = k* \begin{vmatrix} a & b\\ c & d \\ \end{vmatrix} $$
• c2 - row addition $$ \begin{vmatrix} a_{1} & b_{1} \\ c & d \\ \end{vmatrix} + \begin{vmatrix} a_{2} & b_{2} \\ c & d \\ \end{vmatrix} = det\begin{vmatrix} a_{1} + a_{2} & b_{1} + b_{2} \\ c & d \\ \end{vmatrix} $$

From this 3 basic properties you can prove all other properties and formulas.

For instance:

• if 2 rows are equal, the determinant is $0$ (follows directly from b)

• subtracting a multiple of one row from another row does not change the determinant (from the above and c2)

• the determinant of a diagonal matrix is the product of the diagonal elements (from 1 and 3a)

Using the above properties, you can already compute the determinant of a 3x3 matrix - using c1, you can decompose a 3x3 matrix in 27 parts, out of which only 6 have a non-zero determinants: $$ \det(M) =\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = = \begin{vmatrix} a_{11} & 0 & 0 \\ 0 & a_{22} & 0 \\ 0 & 0 & a_{33} \end{vmatrix} + \begin{vmatrix} a_{11} & 0 & 0 \\ 0 & 0 & a_{23} \\ 0 & a_{32} & 0 \end{vmatrix} + \begin{vmatrix} 0 & a_{12} & 0 \\ a_{21} & 0 & 0 \\ 0 & 0 & a_{33} \end{vmatrix} + \begin{vmatrix} 0 & a_{12} & 0 \\ 0 & 0 & a_{23} \\ a_{31} & 0 & 0 \end{vmatrix} + \begin{vmatrix} 0 & 0 & a_{13} \\ a_{21} & 0 & 0 \\ 0 & a_{32} & 0 \end{vmatrix} + \begin{vmatrix} 0 & 0 & a_{13} \\ 0 & a_{12} & 0 \\ a_{31} & 0 & 0 \end{vmatrix} = a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12}. $$ The minus sign correspond to those matrices which requires an odd number of row exchanges to bring them to diagonal form.

For the second part, if you consider that edges of the box are the column vectors of your matrix M, than it's easy to show that the volume of the box comply with the 3 properties of the determinants, which means that the volume is the determinant.