# Cofactor expansion

1. Mar 1, 2008

### fk378

1. The problem statement, all variables and given/known data
Find the determinant of
4...3...-5
5...2...-3
0..-1....2

3. The attempt at a solution
I've tried to get the answer using each of the 3 rows and each time I get a different answer. For the first row I get 14, the second row I get -31, and the third row I get -1. However, I thought that the determinant would be the same value for whichever row (or column) you choose to expand.

2. Mar 1, 2008

### HallsofIvy

Staff Emeritus
Yes, it certainly should be!

Expanding by the first column (since it has only 2 non-zero entries):
$$\left|\begin{array}{ccc}4 & 3 & -5 \\ 6 & 2 & -3 \\ 0 & -1 & 2 \end{array}\right|= 4\left|\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\left|- 5\right|\begin{array}{cc} 3 & -5 \\ -1 & 2\end{array}\right|$$
$$4(4- 3)- 5(6- 5)= 4(1)- 5(1)= -1$$
The determinant is -1. I notice that is what you got using the third row which also has only two non-zero entries. Perhaps it is that third entry that is confusing you.

3. Mar 2, 2008

### fk378

But if you try to work out the determinant using the other rows that do not contain the 0 entry, it does not come out to -1!! Why?

4. Mar 2, 2008

### fk378

Ah, I just realized I wasn't multiplying one of the terms with the cofactor expansion!! I feel silly....thanks for your help though!!