Every determinant is built the same way. Pick one entry from each row, never two from the same column, multiply them, and give the product a sign. Add all such signed products: two for a 2 by 2 matrix, six for 3 by 3, twenty-four for 4 by 4.

Rooks placed on a chessboard so that none attacks another stand one in each row and one in each column. A determinant adds up one signed product for every such placement of rooks on the matrix.

A choice of one entry from each row of an $n\times n$ matrix, no two in one column, is recorded by listing the columns used in rows $1,2,\ldots,n$: a rearrangement $(j_1,j_2,\ldots,j_n)$ of the numbers $1,2,\ldots,n$. There are $n!=1\cdot2\cdots n$ rearrangements. An inversion is a pair of positions where the larger number comes first, and $N(j_1,\ldots,j_n)$ counts them. The determinant adds the product for every rearrangement, with sign $+$ when the number of inversions is even and $-$ when it is odd.

• For $n=2$ the rearrangements are $(1,2)$, with no inversion, and $(2,1)$, with one: $\det A=a_{11}{a}_{22}-a_{12}{a}_{}$, the formula already known.

Take an upper triangular matrix. Row $n$ is zero except in its last column, so a term that is not zero must take $a_{nn}$ there. Row $n-1$ is zero before its own diagonal entry and column $n$ is already used, so it must take , and so on up the matrix. Only the diagonal product survives, and the natural order has no inversions. A lower triangular matrix is the same argument read downwards.

In an upper triangular matrix the last row is zero except in its last column.

So a term that is not zero takes its entry in row $n$ from column $n$.

Row $n-1$ is zero before column $n-1$, and column $n$ is taken. Going up row by row, each row is forced onto the diagonal.

The one surviving rearrangement is the natural order, which has no inversions, so its sign is $+$.

Every other term contains a zero, so the determinant is the product of the diagonal.

Row 1 can use any of the $n$ columns, row 2 any of the $n-1$ left, and so on, down to one column for the last row.

So $2!=2$ terms for a $2\times2$ matrix and $3!=6$ for a $3\times3$ one, as already seen.

Go through the rearrangement number by number and count how many smaller numbers stand to its right. The total is the number of inversions.

Two inversions, an even number, so the sign is $+$.

• A term of an $n\times n$ determinant takes one entry from each row and each column: there are $n!$ of them.
• The columns used, row by row, form a rearrangement; its inversions are the pairs out of order.
• An even number of inversions gives $+$, an odd number $-$.
• The determinant of a triangular matrix is the product of its diagonal.

Great work! You now know how to:

• describe a term of any determinant: one entry per row and per column;
• count inversions and read the sign of a term;
• count the $n!$ terms of an $n\times n$ determinant;
• prove and use the diagonal rule for triangular matrices.