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I am reading Shilov's book linear algebra. He explains how to compute determinants. Basically, for the plus terms you write \begin{equation} x_{a1}x_{b2}x_{c3}x_{d4}x_{e5} x_{f6} \end{equation} and then permute the left side indices, giving \begin{align} &x_{a1}x_{b2}x_{c3}x_{d4}x_{e5} x_{f6}\\ &x_{b1}x_{c2}x_{d3}x_{r4}x_{f5} x_{a6}\\ &x_{c1}x_{d2}x_{e3}x_{f4}x_{a5} x_{b6}\\ &x_{d1}x_{e2}x_{f3}x_{a4}x_{b5} x_{c6}\\ &x_{e1}x_{f2}x_{a3}x_{b4}x_{c5} x_{d6}\\ &x_{f1}x_{a2}x_{b3}x_{c4}x_{d5} x_{e6}\\ \end{align} I just used letters to make it obvious which index was permuting. You should think of it like $a=1,b=2$, etc

For the minus terms you write

\begin{equation} x_{f1}x_{e2}x_{d3}x_{c4}x_{b5} x_{a6} \end{equation}

and do the same thing.

I understood all of that. Made sense.

My Confusion

I want to find the sign of \begin{equation} a_{23}a_{31}a_{42}a_{56}a_{14} a_{65} \end{equation} for this term that appears in a determinant of order $6$.

But if I rearrange to make the right indices in order,

\begin{equation} a_{31}a_{42}a_{23}a_{14} a_{65}a_{56} \end{equation}

you can see the left indices read \begin{equation} 342165 \end{equation} which isn't in order. \begin{equation} 165342 \end{equation}

The above determinant rules only work if the left hand indices are in numerical order like \begin{equation} 432165 \end{equation} My Question:

How can I compute a determinant if the left hand indices aren't in order?

Stan Shunpike
  • 4,921
  • First, there are not 6 permutations but 6! permutations once you include all the shuffling. Secondly, there is no reason to create variables as you have done nor enumerate any of the permutations. All you need to do is count the inversions of the row indices after you've put the columns in order. 342165 has 2 + 2 + 1 + 0 + 1 + 0 = 6 inversions, implying the sign is (-1)^6, or (+). – Darren Ringer Feb 09 '17 at 15:05

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