Proving the formula for finding the determinant of a square matrix.

This is the Definition of Determinant

First of all, they said that a matrix A is invertible (there exists an $n \times n$ square matrix B such that $AB = BA= I_{n}$) if and only if its determinant is nonzero. So, there some kind of scalar value that determines if a matrix if invertible.

Assuming that you have knowledge in row reducing a matrix, we known that there exist a criteria for determining if a matrix is invertible. That criteria is that when row reducing a matrix its entry $a_{nn}$ must not equal zero. That means $a_{nn} \neq 0.$

So let's see the case for a $3 \times 3$ matrix:

$$ \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$

So for row reducing first we have to get zeros in the first column except in the first column, so we can use the row operation of multiplying a row by a scalar. In this case we multiply row 2 and 3 with the scalar $a_{11}$ and then we use the row operation of summing a row multiplied by a scalar, in this case for the second row it will $-a_{21} \cdot R1$ and for third row $-a_{31} \cdot R1$.

$$\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{11}a_{21} & a_{11}a_{22} & a_{11}a_{23} \\ a_{11}a_{31} & a_{11}a_{32} & a_{11}a_{33} \end{bmatrix} \sim \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{11}a_{21} - a_{11}a_{21} & a_{11}a_{22} - a_{12}a_{21} & a_{11}a_{23} - a_{13}a_{21} \\ a_{11}a_{31} - a_{11}a_{31} & a_{11}a_{32} - a_{12}a_{31} & a_{11}a_{33} - a_{13}a_{31} \end{bmatrix} $$

$$ \sim \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ 0 & a_{11}a_{22} - a_{11}a_{21} & a_{11}a_{23} - a_{11}a_{21} \\ 0 & a_{11}a_{32} - a_{11}a_{31} & a_{11}a_{33} - a_{11}a_{31} \end{bmatrix} $$

So we can summarize this steps in a simple algorithm:

This algorithm for getting zeros in the first column can be generalize for an $n \times n$ matrix such that the first step will be:

1. Multiplying all the rows except the first one by the first entry of the first column.

2. Adding to all rows except the first one by the $-a_{i1} \cdot R1.$

So we can apply again this algorithm to the next submatrix $A_{22}$:

After applying the algorithm to the submatrix $A_{22}$ we get th echelon form of the matrix A: $$\sim \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ 0 & a_{11}a_{22} - a_{12}a_{21} & a_{11}a_{23} - a_{13}a_{21} \\ 0 & 0 & (a_{11}a_{33} - a_{11}a_{31})(a_{11}a_{22} - a_{12}a_{21}) - (a_{11}a_{32} - a_{12}a_{31})(a_{11}a_{23} - a_{13}a_{21}) \end{bmatrix} $$

So we known $a_{nn} \neq 0.$, then: $$(a_{11}a_{33} - a_{11}a_{31})(a_{11}a_{22} - a_{12}a_{21}) - (a_{11}a_{32} - a_{12}a_{31})(a_{11}a_{23} - a_{13}a_{21})$$

$$=a_{11}^{2}a_{22}a_{33}-a_{11}a_{12}a_{21}a_{33}-a_{11}a_{13}a_{22}a_{31}+a_{12}a_{13}a_{21}a_{31}-[a_{11}^{2}a_{23}a_{32}-a_{11}a_{13}a_{21}a_{32}-a_{11}a_{12}a_{23}a_{31}+a_{12}a_{13}a_{21}a_{31}]$$

$$=a_{11}[a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}]+a_{12}a_{13}a_{21}a_{31}-a_{12}a_{13}a_{21}a_{31}$$

$$=a_{11}[a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}]+0 \neq 0$$

This is what we call determinant of a $3 \times 3$ matrix. Because it determines if a matrix is invertible or not. $$\Delta = a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}$$

This proccess can be replicated for $n \times n$ matrices.

This algo have a geometrical meaning for $2 \times 2$ matrices and $3 \times 3$ matrices.

Bibliography:

• Lay, David C. Linear Algebra and Its Applications (All chapters but especially chapter 3.1).
• 3Blue1Brown (Youtube Channel)
• http://faculty.fairfield.edu/mdemers/linearalgebra/documents/2019.03.25.detalt.pdf

As an aside, you can apply the cofactor expansion rule to a 2x2 matrix: $$LET:~~\det [x]=x$$ $$\det \left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = a\cdot \det[d]-b\cdot\det[c] = ad-bc$$ Which is exactly the formula you had specified already.

There is nothing to prove actually. What you wrote is the definition of the determinant, and therefore does not need any proof. A definition is just a convention (albeit formal and precise) so that everyone can agree on its meaning.

There are several equivalent ways to define the determinant of a matrix $A \in \mathbb K^{n \times n}$. One could be to define it recursively, starting from the case $n = 1$ or $n = 2$.