Finding the cofactors

1. Nov 11, 2009

thomas49th

Find the cofactors of x,y,z, in the determinate:

| 1 1 1 |
| 2 3 4 |
| x y z |

I can make 3 equations from this by taking the determinate from each line
z - 2y + x
x - 2y + z
z - 6y - x

but how do I solve these? Have I done the right thing? I know they all equal each other

Thanks
Tom?

2. Nov 11, 2009

lanedance

you can take the determinant starting at any column or row, as lonag as you;re careful with negatives

i think the quetsion is asking you to take the determeinant using the bottom row to start off, then it will be something like

det = +-(x(det(2x2)) -y(det(2x2))+ zdet(2x2)

3. Nov 11, 2009

lanedance

the cofactors for x,y,z are the correspsonding 2x2 determinants

4. Nov 11, 2009

lanedance

by the way how do you get these?

5. Nov 12, 2009

HallsofIvy

First, those are not even equations. Second, why? The problem doesn't say anything about "solving", it only asks for cofactors. Do you know what cofactors are?

6. Nov 12, 2009

thomas49th

+ - +
- + -
+ - +

for a 3x3.

The cofactor is taken from the matrix above and placed infront of the orignal matrix element.

in a
1 2 3
4 5 6
7 8 9

the cofactors of row 2 are
-4, +5, -6

is that correct?

EDIT:
I got those equations by taking the determinants of each row. I know that they all equal each other, but I', not sure if that helps!?
You're right HallsOfIvy - those arn't equations, they don't equal anything (except each other). they're just expressions! Silly me!
Thanks
Tom

Last edited: Nov 12, 2009
7. Nov 12, 2009

lanedance

have a look here
http://en.wikipedia.org/wiki/Cofactor_(linear_algebra [Broken])

the cofactor is specific to an element this is what i get

so take the first element of the 2nd row (4) this is row i = 2, column j = 1

the cofactor Cij of 4, is
$$C_{12} = (-1)^{1+2}\left| \begin{matrix} 2&3\\ 8&9 \end{matrix} \right| = (-1)(2.9 - 3.8) = -(18 - 24) = 6$$

Last edited by a moderator: May 4, 2017
8. Nov 12, 2009

thomas49th

ahh once again I've miss understood the workbook. Cofactors are minors with a sign attached, not just the sign!!!

$$A_{31} = 1 A_{32} = -2 A_{33} = 1$$

Thank you very much!