# Number of signed permutation matrices with determinant one

Consider the $$3\times 3$$ permutation matrices with signed ones such that the determinant of the matrix is equal to one. Suppose that it is known that the number of such matrices is $$24$$. What I'm interested in knowing is where my reasoning for the number of such matrices fails.

Let us consider the symmetric group on three letters: $$S_3 = \{(1)(2)(3), (1)(23), (2)(13), (3)(12), (123), (132)\}$$. We can identify each of these permutations with a permutation matrix $$M$$ and then consider the determinant as a multilinear function of the rows of $$M$$. Thus $$\det(M) = \det(r_1, r_2, r_3)$$ and if e.g. $$r_1 = -r_1'$$, then $$\det(M) = -\det(r_1', r_2, r_3)$$. We known that the determinant of an unsigned permutation matrix is equal to the parity of the appropriate permutation. Hence $$\det(M) = \mathrm{sgn}(\sigma_M) * (-1)^n$$, where $$\sigma_M$$ is the appropriate permutation and $$n$$ is the number of negative signs in the matrix.

Therefore I argued that the number of such permutation matrices with determinant one is $$(1 + 3) + (1 + 4) * 3 + (1 + 3) * 2 = 27$$, because the parity of the transpositions is $$-1$$ and the parity of the other three permutations is $$1$$, so that we need to have an odd number of negative ones in each of the tranpositions, i.e. either any of the columns is negative or all of the columns are negative, and an even number of negative ones (in total $$\mathrm{nCr}(3, 2) = 3$$) for the other three permutations.

But evidently this reasoning is not correct as the total number of such matrices is 24. So where did I go wrong?

• The $1+4$ should be $1+3$. I'm assuming this is meant to be the number of ways of choosing an odd number of negatives ones. Commented Mar 3, 2022 at 9:29
• A nicer way of counting is just to observe that the group without the determinant condition has order $2^3 \cdot 6 = 48$, and the determinant is a surjective homomorphism to a group of order $2$, so its kernel has order $24$. Commented Mar 3, 2022 at 9:30
• @SeanEberhard But how is it $1 + 3$? We can choose any of the rows to be negative (3 choices) or set all of the rows to negative (one choice), giving us in total four choices in addition to the choice of not setting any of the rows to be negative. Commented Mar 3, 2022 at 10:28
• If you set none of the rows to be negative you have an even number of -1's, which is not allowed. Commented Mar 3, 2022 at 10:29

All of the factors should be $$1+3$$. For an even number of negative entries, those are the counts for $$0$$ and $$2$$ negative entries, respectively, and for an odd number of negative entries, they’re the counts for $$3$$ and $$1$$ negative entries, respectively.