Question

Linear Algebra - Determinants: I'm stumped. I need to show that for an nth order determinant, by expansion, that there will be n! multiplications. I'm not able to show the n! multiplications, however. Any ideas?

Answer 1

use induction.
the number of multiplies for a 2x2 is easy to count (hint: it's 2)
assume the number of multiplies for an n x n matrix is n!
now examine a (n+1) x (n+1) matrix. How do you find its determinant?
Choose the 1st row. and use cofactors.

Answer 2

Right, so the idea here is to fix a row in the

\[\sum_{i=1}^{n} a _{ij} C _{ij}\]
with k=1, for instance. I'm missing something in the expansion that is not giving me the n! multiplications. I'm just not sure where that is. Likely my expansion is incorrect somewhere, though. Obviously n=2 gives 2 multiplication, n=3 gives me 3 multiplications, not 6. If I try the expanded form with

\[C _{ij}=(-1)^{i+j}M _{ij}\]

I can get 6 total multiplications for n=3, and yet it falls apart at n=4.

Answer 3

Potentially if I expand out in matrix Minors form? So an n=3 order matrix that would be

\[\left[\begin{matrix}a & b & c \\ d & e &f \\ g & h & i\end{matrix}\right]\]

By expansion would be

Det = \[a \left[\begin{matrix}e & f \\ h & i\end{matrix}\right] - d \left[\begin{matrix}b & c \\ h & i\end{matrix}\right] + g\left[\begin{matrix}b & c \\ e & f\end{matrix}\right]\]

But that doesn't yield me 6 multiplications. I count 12.

Answer 4

I got it. My definition of a multiplication was skewed. There are n! multiplication groups. I took n=3 matrix and counted each multiplication of terms, they want multiplication groups. Silly.