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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?

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  • $\begingroup$ It depends on your familiarity with complicated algebra expressions but I think they don’t teach the proof in high school class $\endgroup$
    – acat3
    Commented Aug 16, 2021 at 6:13
  • $\begingroup$ @Rezha If I want to understand it, are there any resources for further reading? Online would be very appreciated. $\endgroup$
    – Mick_Mick
    Commented Aug 16, 2021 at 6:20
  • $\begingroup$ 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 $\endgroup$
    – acat3
    Commented Aug 16, 2021 at 9:15

2 Answers 2

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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..

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YouTube - Proof of the Cofactor Expansion Theorem – Brian V

This YouTube video might help you.

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