List explicitly the six permutations of degree 3, state which are odd and which are even, and use this to give the complete formula (5-15) for the determinant of a \(3 \times 3\) matrix.
Solution
The six permutations of \(\{1,2,3\}\) are \[ \mathrm{id},\quad (123),\quad(132),\quad(12),\quad(13),\quad(23). \] The identity and the two \(3\)-cycles are even; the three transpositions are odd. Therefore, for \(A=(A_{ij})\), the permutation definition of determinant gives \[ \begin{aligned} \det A={}&A_{11}A_{22}A_{33}+A_{12}A_{23}A_{31}+A_{13}A_{21}A_{32}\\ &-A_{12}A_{21}A_{33}-A_{13}A_{22}A_{31}-A_{11}A_{23}A_{32}. \end{aligned} \] These are exactly the three even-permutation terms minus the three odd-permutation terms.