I have read the proof for finding the determinant of a $2 \times 2$ matrix. It makes sense, since for a matrix \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} $(ad-bc)$ must be non-zero for the inverse of the matrix to exist. So it is logical that $(ad-bc)$ is the determinant.
However when it comes to a $3 \times 3$ matrix, all the sources that I have read purely state that the determinant of a $3 \times 3$ matrix defined as a formula (omitted here, basically it's summing up the entry of a row/column * determinant of a $2 \times 2$ matrix). However, unlike the $2 \times 2$ matrix determinant formula, no proof is given.
Similarly, the formula for the determinant of an $n \times n$ matrix is not given in my textbook. Unfortunately, I can't seem to find a proof that I could comprehend on the internet. It would be great if someone can give me a proof of the formula for finding the determinant of an $n \times n$ matrix.
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$\begingroup$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:
Multiplying all the rows except the first one by the first entry of the first column.
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)
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 $n = 2$.
$\endgroup$ 2 $\begingroup$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.
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