Starting with the Leibniz formula for the determinant, I wish to derive the Laplace (Cofactor) Expansion. At the risk of being overly verbose, please see the proof here. Now I understand the idea of the proof. The difficulty I have is in how they managed to obtain that $$\tau\,=\left(n,\ n-1,\ \ldots,\ i\right)\circ\sigma^\prime\circ\left(j,\ j+1,\ \ldots,\ n\right)$$ I can see how $\sigma$ (and hence $\sigma^\prime$) is obtained from $\tau$ by shifting the indexes. However, I am unable to make an intuitive derivation of the above formula. If anyone can help shed some light that would be greatly appreciated.

  • $\begingroup$ Alternatively and equivalently, $det(A) = \Sigma_{i_1,i_2, \dots , i_n}\epsilon_{i_1,i_2, \dots , i_n}A_{1i_1}A_{2i_2} \cdots A_{ni_n}$. The cofactor expansion follows from expanding a particular index of the sum and taking care to sort through the sign-flips from permutations of the indices of the completely antisymmetric symbol. I guess this may not be terribly useful to your way of thinking about the derivation. $\endgroup$ – James S. Cook Aug 13 '12 at 4:43

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