What is the best method to get a matrix into upper triangular form?

[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 I am struggling. I get as far as:

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

Now I am 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.

 

[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 :)

 

[Firepanda]

-12 is my answer if that looks correct.

 

[CompuChip]

next i was going to use row3 and add on 2 lots of row1

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.

 

[Firepanda]

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

 

[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?

 

[Firepanda]

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

 

[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

 

[cristo]

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

Maple tells me the answer's zero. We can't really help you unless you show every row operation.

 

[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
:)

 

[Firepanda]

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

 

[cristo]

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

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

 

[Firepanda]

Ah dammit.

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

 

[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

 

[CompuChip]

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