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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    That formula is the _definition_ of the determinant! You can't "prove" a definition; we can define any word to mean anything we like... – David C. Ullrich Jul 08 '15 at 17:33
  • You might want to consult the wikipedia on [Laplace expansion](https://en.wikipedia.org/wiki/Laplace_expansion). – user251257 Jul 08 '15 at 17:34
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    read this https://en.wikipedia.org/wiki/Laplace_expansion the function determinant can be defined by recursion, the base case for $n=2$, and the Laplace explansion can be used for the $n>2$ term. – L F Jul 08 '15 at 17:35
  • @DavidC.Ullrich you don't know which definition he exactly knows ;) – user251257 Jul 08 '15 at 17:36
  • @user251257 I don't "know" with mathematical certainty what definition is in force, but I'd bet a lot of money that the formula the OP is asking about is in fact the definition in the book for the course. – David C. Ullrich Jul 08 '15 at 17:42
  • hmm, I understand it's a definition. But why exactly is the definition a suitable one? How is it defined in such a way? (I might be making a fool of myself, but I think a definition has it's origins and behind-the-scene logic) Thanks for your suggestions! Unfortunately, I can't really understand the proof on wikipedia. Is there a simpler way to prove the idea? – Nick123 Jul 08 '15 at 17:47
  • Unfortunately I don't have the answer, but regardless of how one defines the determinant, I think the questioner wants proof of the equivalence between the permutation formula, the algorithm using submatrices and the fact that the determinant is the only alternating n-linear form etc etc And so do I ! – James Well Jul 31 '18 at 22:49

4 Answers4


This is the Definition of Determinant

enter image description here

Sudo su
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    link of the book https://www.dropbox.com/s/ydt3aoa8ejpekbv/Bookfile.pdf?dl=0 – Sudo su Jul 08 '15 at 17:46
  • Unfortunately in most cases it is painful to calculate. – Vim Jul 08 '15 at 18:38
  • @Sudosu, I know it's been quite a while since you answered this, but the linkyou've posted above seems to be broken. Do you by any chance still have a good link to the same? Thanks : ) – Gaurav Chandan Jul 08 '21 at 10:46

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: enter image description here

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 $A_{22}$: enter image description here

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: $$(a_{11}a_{33} - a_{11}a_{31})(a_{11}a_{22} - a_{12}a_{21}) - (a_{11}a_{32} - a_{12}a_{31})(a_{11}a_{23} - a_{13}a_{21})$$



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

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.


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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 $n = 2$.

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  • Is there a reason why it's defined this way? – Nick123 Jul 08 '15 at 18:05
  • @Nick123 Yes, because the properties of the determinant make it a unique function that maps $N$ vectors to a real number. For more intuition, check this brilliant answer that explains it better than I possibly could: http://math.stackexchange.com/a/669/2582 – rubik Jul 08 '15 at 19:18

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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