# A simple question about determinant

Let $a_i=[a_{i1},a_{i2},\ldots,a_{in}]\in \mathbb{R}^n$, for $i=1,\ldots,n-1$. How to prove that

$$\sum_{i_1,\ldots,i_{n-1}=1}^n \varepsilon_{i,i_1,\ldots,i_{n-1}} a_{1,i_1} a_{2,i_2}\cdots a_{n-1, i_{n-1}}= (-1)^{1+i} \det \left [ \begin{array}{rrrrr} a_{11} & \ldots &\hat{a_{1i}} & \ldots & a_{1n}\\ a_{21} & \ldots & \hat{a_{i2}} & \ldots & a_{2n} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{n-1,1}& \ldots & \hat{a_{n-1,i}} & \ldots & a_{n-1,n} \end{array} \right ],$$

for $i=1,\ldots,n$, where $\varepsilon_{i,i_1,\ldots,i_{n-1}}=1$ or $-1$ or $0$ depending on whether $(i,i_1,\ldots,i_{n-1})$ is an even permutation or an odd permutation or it is not a permutation of numbers $1,\ldots,n$.

(Symbol $\hat{a_{ij}}$ means that $a_{ij}$ is omitted.)

Thanks.

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The RHS is the determinant of $\pmatrix{0&\ldots&0&1&0&\ldots& 0 \\\ a_{11}&\ldots& a_{1,i-1}&a_{1,i}&a_{1,i+1}&\ldots &a_{1,n}\\\ a_{21}&\ldots &a_{2,i-1}&a_{2,i}&a_{2,i+1}&\ldots &a_{2,n}\\\ \vdots&\ldots&\vdots& \vdots&\vdots&\cdots &\vdots\\\ a_{n-1,1}&\ldots& a_{n-1,i-1}&a_{n-1,i}&a_{n-1,i+1}&\ldots &a_{n-1,n} }$, and putting $b_{ij}=a_{ij}$ if $i\leq n-1$, $a_{nj}=\delta_{ij}$ it's equal to $\sum_{\sigma\in\Sigma_n}\varepsilon(\sigma)\prod_{j=1}^nb_{j,\sigma(j)}$. In fact you have to consider the permutations which fix $i$, which will give you the result.

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Maybe the "zero-one" row should be in the first row? Thanks a lot for answer. –  Richard Mar 3 '12 at 18:15
You are right, it should indeed the first. Thanks for pointing this out. –  Davide Giraudo Mar 3 '12 at 18:24