number of diagonal components in term of determinant?

Question

In linear algebra a determinant of an nxn matrix is the sum n! terms each of which is the product of n components of the matrix. One of the terms contain all n diagonal components. It appears that none contain exactly n-1 diagonal components. (I tested it for 2x2,3x3, 4x4 matrices.) There are many with fewer diagonal components. Is it true that none contain exactly n-1 diagonal components?

Answer 1

Yes, that is correct. If your matrix is $(a_{ij})_{1\leqslant i,j\leqslant n}$, then each term of the determinant is $\pm\prod_{i=1}^na_{i\sigma(i)}$, shere $\sigma$ is some permutation of $\{1,2,\ldots,n\}$. But then either $\sigma$ is the identity permutation, in which case you get the product of all elements of the main diagonal, or there are at least two distinct elements $j,k\in\{1,2,\ldots,n\}$ with $\sigma(j)\ne j$ and $\sigma(k)\ne k$. Then $\prod_{i=1}^na_{i\sigma(i)}$ has at most $n-2$ entries from the main diagonal.