TITLE: How to deduce that there are $(n^3+2n-3)/3$ multiplications for the determinant evaluation?
QUESTION [6 upvotes]: In Friedberg's Linear Algebra, the author points out that the evaluation of the determinant of an $n\times n$ matrix by cofactor expansion along any row requires over $n!$ multiplications, whereas evaluating the determinant of an $n\times n$ matrix by elementary row operations can be shown to require only $(n^3+2n-3)/3$ multiplications.
I cannot even figure out the case when $n=2$. The key point, I think, is the exact algorithm of evaluating by elementary row operations. But I temporarily have no idea how to go on. 
Here is my question:
How can I deduce the number $(n^3+2n-3)/3$?

REPLY [6 votes]: Assuming the $(1,1)$ entry is non-zero, you have to do $n$ multiplications to use the first row to wipe out the $(2,1)$ element, another $n$ to wipe out $(3,1)$, etc.: all told, $n(n-1)$ operations to wipe out the first column. Then you have to do $((n-1)^3+2(n-1)-3)/3$ multiplications to get the determinant of the $(n-1)\times (n-1)$ matrix, and one more multiplication to finish things off. Does it add up right?