Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)


share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.