# Proof of the Laplace Expansion? [closed]

I just learned about Laplace Expansion for determinant calculation in high-school, they taught me how to calculate minors cofactors and everything but they did not include the proof in the book. I have tried watching YouTube videos and there is a proof on Wikipedia which I cannot understand anything.

My teacher says she doesn't know the proof. Is the proof too hard for a high school student?

• It depends on your familiarity with complicated algebra expressions but I think they don’t teach the proof in high school class Commented Aug 16, 2021 at 6:13
• @Rezha If I want to understand it, are there any resources for further reading? Online would be very appreciated. Commented Aug 16, 2021 at 6:20
• I posted an overview of the proof. Once again, this is not normally taught in high school and even your teacher doesn’t know the proof. However, I hope it’s enough to give you a sense of why the expression is correct. If you are still curious, you need to start reading books about Linear Algebra. Any book should be ok although I prefer books from Springer Commented Aug 16, 2021 at 9:15

First you need to understand odd and even permutation. Define $$\sigma$$ as the permutation of the first $$n$$ positive integers:

$$\sigma=(\sigma_{1}, \sigma_{2},…,\sigma_{n})$$

Define $$s_{i}$$ as the number of $$j >i$$ such that $$\sigma_{i}>\sigma_{j}$$. If $$\sum_{i=1}^{n}{s_{i}}$$ is odd we call $$\sigma$$ an odd permutation and even permutation otherwise. Then we define the $$sgn$$ function below:

$$sgn(\sigma)= \begin{cases} \phantom{-}1\phantom{xx}\text{even}\phantom{x}\sigma\\ -1\phantom{xx}\text{odd}\phantom{x}\sigma \end{cases}$$

We also define a special permutations; $$\tau^{j}$$ as the permutation of the first $$n$$ positive integers except $$j$$. Notice the following:

\begin{align} &\text{if}\\ &\phantom{xx}\sigma^{*}=(j,\tau^{j}_{1},\tau^{j}_{2},…,\tau^{j}_{n-1})\\ \\ &\text{then}\\ &\phantom{xx}sgn(\sigma^{*})=\left(-1\right)^{j-1}sgn(\tau^{j}) \end{align}

Now we move on to the definition of determinant using Leibniz formula:

$$|A|=\sum_{all\phantom{x}\sigma}\left(sgn(\sigma)\prod_{i=1}^{n}{a_{i,\sigma_{i}}}\right)$$

We can rewrite this equation as the following

\begin{align} |A|&=\sum_{i=1}^{n}{\left(a_{1,i}\sum_{all\phantom{x}\sigma^{*}}{\left(sgn(\sigma^{*})\prod_{j=1}^{n-1}{a_{(j+1),\tau^{i}_{j}}}\right)}\right)}\\ \\ &= \sum_{i=1}^{n}{\left((-1)^{i-1}\phantom{x}a_{1,i}\sum_{all\phantom{x}\tau^{i}}{\left(sgn(\tau^{i})\prod_{j=1}^{n-1}{a_{(j+1),\tau^{i}_{j}}}\right)}\right)}\\ \\ &= \sum_{i=1}^{n}{\left((-1)^{i-1}\phantom{x}a_{1,i}|A_{(1)(i)}\phantom{.}|\right)} \end{align}

Above, $$A_{(1)(i)}$$ is the matrice $$A$$ with the first row and $$i$$ -th column deleted. I think the last part is familiar with you since you have learned about minor, cofactor, etc..

YouTube - Proof of the Cofactor Expansion Theorem – Brian V