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I want to inductively proof that the denominator of the general solutions of an $n \times n $ system are given by the Leibnitz formula for the determinant \begin{align*} \sum_{\sigma \in S_{n}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} a_{i, \sigma(i)} .\end{align*}

For the induction step, suppose our matrix $\mathbf{A}\in \mathbb{R}^{n+1, n+1} $ is given by $( \mathbf{A})_{i, j} = a_{i, j}$. My idea is now to eliminate the first column of the matrix (normal gaussian elimination) and consider the $(n\times n)$ submatrix $(\mathbf{A})_{1:, 1:} = \widetilde{\mathbf{A}}$ with \begin{align*} (\widetilde{\mathbf{A}})_{i, j} = \frac{a_{i+1, j+1}\cdot a_{1, 1}- a_{1, j+1}\cdot a_{i+1, 1}}{a_{1, 1}} .\end{align*} To this matrix we can apply the formula we assume to be true, so what I want to show is \begin{align*} a_{1, 1}\sum_{\sigma \in S_{n}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} (\widetilde{\mathbf{A}})_{i, \sigma(i)} = \sum_{\sigma \in S_{n+1}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n+1} a_{i, \sigma(i)} .\end{align*} (The additional factor $a_{1, 1}$ comes from the fact that for getting the denominator of the general solution we have to bring it to the other site and there it will cancel out, but this we don't consider when applying the formula directly to the submatrix). However, I didn't come very far in my proof \begin{align*} a_{1, 1}\sum_{\sigma \in S_{n}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} (\widetilde{\mathbf{A}})_{i, \sigma(i)} &= a_{1, 1}\sum_{\sigma \in P_{n+1}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} \left( \frac{a_{i+1, \sigma(i+1)}\cdot a_{1, 1}- a_{1, \sigma(i+1)}\cdot a_{i+1, 1}}{a_{1, 1}} \right) \\[5pt] &= a_{1, 1}\sum_{\sigma \in P_{n+1}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 2}^{n+1} \left( \frac{a_{i, \sigma(i)}\cdot a_{1, 1}- a_{1, \sigma(i)}\cdot a_{i, 1}}{a_{1, 1}} \right) \\[5pt] &= \sum_{\sigma \in P_{n+1}}^{} \frac{\text{sgn}\left(\sigma\right)}{(a_{1, 1})^{n-1}} \prod_{i = 2}^{n+1} \left( {a_{i, \sigma(i)}\cdot a_{1, 1}- a_{1, \sigma(i)}\cdot a_{i, 1}} \right) \end{align*} Could someone help me with the proof?

Edit: $P_{n+1}$ denotes the permutations of the elements $\{2, 3, \ldots, n+1\}$. Also, in the comments the Neville algorithm got mentioned, but I don't see the connection yet.

Rhi
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  • Related https://en.wikipedia.org/wiki/Neville%27s_algorithm – Miss and Mister cassoulet char Jan 02 '22 at 16:14
  • @ErikSatie could you maybe elaborate? I don't immediately see the connection. – Rhi Jan 02 '22 at 16:20
  • Are you trying to prove Cramer's rule (https://en.wikipedia.org/wiki/Cramer%27s_rule#General_case)? – Mason Jan 02 '22 at 22:56
  • @Mason not really, I'm trying to proof Leibniz's rule (but not from the constructive point of view where you derive it using the properties) – Rhi Jan 03 '22 at 10:19
  • @Rhi What exactly are you trying to prove? The thing about the determinant being in the denominator of the solutions sounds like Cramer's rule. – Mason Jan 04 '22 at 20:28
  • @Mason No, the determinant of the general solution (i.e. if you solve the system for $x_1$, $x_2$, ...) is given by Leibniz's rule, and this is what I'm trying to prove. – Rhi Jan 05 '22 at 12:06
  • @Rhi Are you trying to prove that $\det(A) = \sum_{\sigma \in S_n}\text{sgn}(\sigma)a_{1, \sigma(1)}\dots a_{n, \sigma(n)}$? If so, how are you defining $\det$? – Mason Jan 07 '22 at 18:42
  • @Mason Yes, but I pretend to not know the determinant yet, I only consider the denominator of the general solution. – Rhi Jan 12 '22 at 00:03
  • @Rhi How can you pretend to not know the determinant, yet you know Leibniz formula for the determinant?! And what do you mean exactly by "denominator of the general solution"? – Mason Jan 12 '22 at 01:44
  • @Mason If you have a general system of $n$ linear equations and solve this system, int the solution vector $\mathbf{x}$, every entry will have the same denominator (assuming sufficient conditions so that this denominator isn't zero). Due to observations one can conjecture the Leibniz formula for the determinant yields this denominator (hence the denominator is what we call the determinant today). So essentially I'm trying to derive the determinant in a different way. – Rhi Jan 12 '22 at 16:06

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