# Homework Help: Determinants from any row or column

1. Jul 6, 2011

### Fan de Douze

I'm having a problem with this rule in general. Apparently one can calculate the determinant by multiplying the cofactors and entries of any row or any column of a matrix. I have a negative that pops up. I'll take a 3X3 matrix for simplicity.

A=
|a b c|
|d e f|
|g h i|

calculating from the top row we have:
det(A)=a(ei-fh)-b(di-fg)+c(dh-eg) = aei-afh-bdi+bfg+cdh-ceg
calculating from the middle row we have
det(A)=d(bi-ch)-e(ai-cg)+f(ah-bg) = -aei+afh+bdi-bfg-cdh+ceg

The determinant seems to change by a factor of negative 1. This also seems to make sense from row operations. If for example, I swap the first and the second row to yield a matrix B, Then det(B)=-det(A). Yet if I calculate using entries and cofactors corresponding to the second row of B and the first row of A, the cofactor expansion of the two determinants will be identical. Though I'm told that we can start from any row or column, what did I miss? Thanks all.

2. Jul 6, 2011

### HallsofIvy

That's because you are doing the second one incorrectly. The rule is that if you have $a_{ij}$ times its cofactor, you must multiply by $(-1)^{i+j}$. The simplest way to get that factor is to start at the top left and move by horizontally or vertically, alternating "+" and "-".

So for your first calculation, "expanding on the first row", you should have "+, -, +" ($a_{11}$: $(-1)^{1+1}= 1$, $a_{12}$: $(-1)^{1+2}= -1$, $a_{13}$: $(-1)^{1+ 3}= 1$) as you have. But "expanding on the second row", it should be "-, + , -" ($a_{21}$: $(-1)^{2+1}= -1$, $a_{22}$: $(-1)^{2+2}= 1$, $a_{23}$: $(-1)^{2+3}= -1$) which changes the sign.

Last edited by a moderator: Jul 6, 2011
3. Jul 6, 2011

### Fan de Douze

Thank you, perfect, so I was using the minor of each entry which doesn't include the (-1)^(i+j), which is necessary for the cofactor because Cij=(-1)(i+j)Mij. Thank you.