Cofactor (4x4 matrix)

1. Oct 19, 2009

shiri

For the matrix A given below, find C44 , where Cij is the i,j cofactor of A.

A=

| -5 +4 +5 -3 |
| +2 -1 +5 +4 |
| -5 -1 +2 -3 |
| +4 -2 +4 -4 |

So this is how I answer this question:

= (4,4) cofactor of A

=
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| -5 -1 +2 |

=
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| +5 +1 -2 |
R3*(-1)

=
| +0 +5 +3 |
| +2 -1 +5 |
| +5 +1 -2 |

=
| -1 +5 | *(+)(+)
| +1 -2 |
(1,1) cofactor of A

=(+)(+)[(-1*-2)-(1*5)]

=(+)(+)(2-5)

=-3

2. Oct 19, 2009

tiny-tim

Hi shiri!
No, that's the cofactor of the +0, and you get the determinant by multiplying +0 times its cofactor (and then adding the same for +5 and +3).

If you're determined to save effort by getting down to a 2x2 determinant, you need another 0.

I'd have started differently, and used one of the original -1s to get rid of the other -1 and the 4.

3. Oct 19, 2009

shiri

Hi

… =
| +0 +5 +3 |
| +2 -1 +5 |*(+)
| +5 +1 -2 |

=
| +0 +5 +3 |
| +2 -1 +5 |*(+)
| +0 +7 -29 |
2R3 - 5R2

=
| +5 +3 | *(+)(-)
| +7 -29 |
(2,1) cofactor of A

=(+)(-)[(+5*-29)-(+7*+3)]

=(+)(-)[(-145)-(21)]

C44 = 166

4. Oct 19, 2009

tiny-tim

hmm … you've made two mistakes, and still got the right answer!

i] you can add as many copies as you like of one row to another row,

but you can't add copies of a row to itself … if you do that, you multiply the determinant (in this case by 2, when you added R3 - 5R2 to the original R3)

ii] you should have multiplied the cofactor, 166, by the 2 that it's the cofactor of.

Try again.

5. Oct 19, 2009

shiri

So the C44 = 332?

6. Oct 19, 2009

tiny-tim

No!!

I said 166 was right …

you made two mistakes, one doubled the answer, and the other halved it again.

Do you understand why your 2R3 - 5R2 was wrong?​

7. Oct 19, 2009

shiri

Strange. When I typed 166 in the online assignment. It came out wrong. Do you think the answer was suppose to be -166, not 166?

8. Oct 20, 2009

tiny-tim

(just got up :zzz: …)
oops! yes, I didn't notice that you made a third mistake (I thought that the two matrices were the same ) …
when you multiply a row by minus one, you multiply the whole determinant by minus one!!

As I said before, you can only add multiples of one row to another row!

Now try the easier way I suggested before: use one of the original -1s to get rid of the other -1 and the 4.

9. Oct 20, 2009

shiri

So it becomes like this:

| -5 +4 +5 |
| +2 -1 +5 | * (+)
| -5 -1 +2 |

=
| -5 +5 | * (+) (+)
| -5 +2 |

=(+)(+)[(-5*+2)-(-5*+5)]

=(+)(+)[-10-(-25)]

=(+)(+)[-10+25]

=(+)(+)[15]

So is that means, C44 = 15?

10. Oct 20, 2009

tiny-tim

That's bizarre!

All you've done is calculate the cofactor of (an arbitrary) -1.

Add multiples of the 2nd row to the 1st and 3rd rows so as to turn the -1 and the 4 into 0s …

then!!! you can use the cofactor of the remaining -1 !

11. Oct 20, 2009

shiri

OK. I hope I know what I'm doing.

| -5 +4 +5 |
| +2 -1 +5 | * (+)
| -5 -1 +2 |

=
| -5 +4 +5 |
| +2 -1 +5 | * (+)
| +7 0 +3 |
R3(-1) + R2

=
| +3 +0 +25 |
| +2 -1 +5 | * (+)
| +7 +0 +3 |
R1 + 4R2

=
| +3 +25 | *(+)(+)
| +7 +3 |

=(+)(+)[(+3*+3)-(+7*+25)

=(+)(+)[+9-175]

=-166

C44 = -166

Am I doing this right, tiny-tim?

12. Oct 20, 2009

tiny-tim

Yes!!

Do you see why this works? …

the determinant is the sum of each element in one row (or one column) multiplied by its cofactor …

so you made a row that was all 0s except for one element …

so the determinant is that element multiplied by its cofactor!

(and you chose -1 because keeping the numbers small, and preferably ±1, makes the arithmetic a lot easier than eg 2 and 5 )

13. Oct 20, 2009

shiri

Great. Thank You! You're very helpful.

Oh, one more thing. Can you take a look at this question [https://www.physicsforums.com/showthread.php?t=346994 Matrix Determinant]

I just want to know whether I answer this question correctly. Thanks