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I am trying to get my head around the Leibniz formula for this $3 \times 3$ determinant. I know that

So for $n=3$ we get $$\det(A) = a_{1,1}a_{2,2}a_{3,3} - a_{1,1}a_{2,3}a_{3,2} - a_{1,3}a_{2,2}a_{3,1} - a_{1,2}a_{2,1}a_{3,3} + a_{1,2}a_{2,3}a_{3,1} + a_{1,3}a_{2,1}a_{3,2}$$

But I don't know how we work out the signs in each case above? Could someone show me how? I know the sign is $+$ when the permutation is even but how is $a_{1,1}a_{2,2}a_{3,3}$ even but $a_{1,2}a_{2,1}a_{3,3}$ is odd?

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  • $\begingroup$ both of them have negative signs in what you wrote. $\endgroup$
    – Phicar
    Nov 29, 2020 at 15:26
  • $\begingroup$ Did you even look at the indices properly? There is a minus sign in front of $a_{13}a_{22}a_{31}$. $\endgroup$ Nov 29, 2020 at 15:27
  • $\begingroup$ Sorry my mistake, corrected now $\endgroup$
    – user635953
    Nov 29, 2020 at 15:30
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    $\begingroup$ There are many different ways to find the sign of a permutation. The formular $\prod_{i<j} \frac{i-j}{\sigma(i)-\sigma(j)}$, counting inversions, using cycle length in the disjoint cycle decomposition, counting how many transpositions you need to decompose $\sigma$ as a product of such, … Pick the one you like best! $\endgroup$
    – Christoph
    Nov 29, 2020 at 15:34

1 Answer 1

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An alternative odd/even property is the parity of the number of inversions. An inversion of a permutation $\sigma$ is defined to be the set $$Inv(\sigma)=\{(i,j):i<j\text{ and }\sigma_i>\sigma _j\}.$$ So, in your case $Inv(321)=\{(1,2),(1,3),(2,3)\}$ because $3>1$ and $3>2$ and $2>1$ is odd because it has $3$ inversions and $Inv(213)=\{(1,2)\}$ (the only one is $2>1$) also has $1$ so is odd. In general, the sign of a permutation $\sigma$ is $(-1)^{|Inv(\sigma)|}.$

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  • $\begingroup$ I see! Thank you! $\endgroup$
    – user635953
    Nov 29, 2020 at 15:31
  • $\begingroup$ @JamesO'dare You are welcome! $\endgroup$
    – Phicar
    Nov 29, 2020 at 15:32

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