# Homework Help: Determinant of Matrices

1. Dec 2, 2007

### Firepanda

[SOLVED] Determinant of Matrices

Find the det. of the matrix:

1 3 -2 2
3 0 -1 4
1 -3 4 2
2 3 -3 3

Using gaussian elimination to get it into upper triangular form (which I think is easiest) is where im struggling. I get as far as:

1 3 -2 2
0 -9 5 -2
0 -6 6 0
0 -3 0 -1

Now im trying to get everythingbelow the diagonal all zeros, next i was going to use row3 and add on 2 lots of row1 to get a 0 for the -6, but it also removes one of my zero values and replaces it as a 2.

I guess the method I am using is wrong to get it into upper traingular form, can anyone tell me where I am going wrong? Perhaps I should be using column operations instead.

2. Dec 2, 2007

### Firepanda

Actually perhaps I'm being stupid and doing this completely wrong, I think I can use cofactors instead, I don't know where I got the other method from, I guess it was for something else to do with matrices :)

3. Dec 2, 2007

### Firepanda

-12 is my answer if that looks correct.

4. Dec 2, 2007

### CompuChip

That's not going to work (as you already found out), instead add multiples of the second row to the third and fourth one (e.g. add the second to the first -2/3 times)

By the way, the answer of -12 is not correct.

5. Dec 2, 2007

### Firepanda

awww, was the method of cofactors bad? :P cos I only did it with 3x3 matirces in practise.

Last edited: Dec 2, 2007
6. Dec 2, 2007

### CompuChip

Just looked up "cofactor" on Wikipedia -- obviously I know the method but I never heard that name
That method will also work for any matrix (though the larger the matrices get, the more work it will be), so you probably just made a calculation error.
Can you post a calculation?

7. Dec 2, 2007

### Firepanda

actually I just got the answer using my original method as -12 again, are you sure it's not correct?

8. Dec 2, 2007

### Firepanda

my end matrix for the original method is:

1 3 -2 2
0 -9 5 -2
0 0 8/3 4/3
0 0 0 0.5

and 1*-9*(8/3)*0.5 is -12 :P

9. Dec 2, 2007

### cristo

Staff Emeritus
Maple tells me the answer's zero. We cant really help you unless you show every row operation.

10. Dec 2, 2007

### Firepanda

Well in my original method my row operations were:

R2 - 3R1 and R3 - R1 and R4 - 2R1

to give

1 3 -2 2
0 -9 5 -2
0 -6 6 0
0 -3 0 -1

then, R3 - (2/3)R2 and R4 - (1/3)R2

to give

1 3 -2 2
0 -9 5 -2
0 0 (8/3) (4/3)
0 0 (-5/3) (-1/3)

then finally R4 + (5/8)R3

to give

1 3 -2 2
0 -9 5 -2
0 0 8/3 4/3
0 0 0 0.5
:)

11. Dec 3, 2007

### Firepanda

It all seems ok to me, considering how I got the same answer for both methods, unless the methods were wrong.

12. Dec 3, 2007

### cristo

Staff Emeritus
The last row is wrong; that is, check "R4-2R1."

13. Dec 3, 2007

### Firepanda

Ah dammit.

Still weird I ended up with the same answer for both :P

14. Dec 3, 2007

### Firepanda

Nono, w8 w8, the start matrix i wrote down here wasn't correct :P it should have been:

1 3 -2 2
3 0 -1 4
1 -3 4 2
2 3 -4 3

=P I hope I'm correct now, sorry lol, guess i was typing out the numbers too quick

15. Dec 3, 2007

### CompuChip

That has determinant -12 indeed, so you were correct after all
It's often in the little steps