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So the third row would be [1, -1, 0, 2]

Second term being -1

So do I just use -1 times the rest of the matrix?

I guess you can take the determinant?

Then multiply by -1

Take the determinant of the original matrix and times that by -1?

whats the definition of "term" that your professor uses ?

does the determinant has \(4!=24\) terms or just \(4\) terms according to your prof ?

I think she'd say that that'd be just 4 terms.

**
cofactor of \(A_{32}\)= \((-1)^{3+2}\begin{vmatrix}-1&-1&-1\\-1&2&0\\-1&-1&-1\end{vmatrix}\)

Okay, I am obviously not understanding this question because the answer I got was 0...

Right, I think your prof wants the first term out of those 24 terms in the determinant

the answer is either 0 or 2
0 if you consider 4 terms in the determinant
2 if you consider 24 terms

yahoo answers is not that reliable..

Aha, I know, I was just trying to work out how they managed to get -20 that's all

they got -40 right ?

Yeah.

remember how to find cofactor of a term ?

Yes, but it's a bit hard to write it all on here.

removing 3rd row and 2nd column,
[3,2,-1]
[-2,1,2]
[-1,2,-1]

And then multiplying all that by -2.

What I don't understand is how they ended up getting (-2) x (-20) to get 40.

before that, multiply the determinant by \((-1)^{3+2}\) to get the cofactor

what do you get for determinant of
[3,2,-1]
[-2,1,2]
[-1,2,-1]

Ohh... My bad..

lets work it step by step

how many terms are there in the determinant of a 2x2 matrix ?

\[ \begin{vmatrix}a&b\\c&d\\\end{vmatrix} = ad-bc\]
two terms, right ?

Yes.

how about the determinant of a \(3\times 3\) matrix, how many terms ?