Finding determinant with upper triangular matrix

[theRukus]

So. I've been told by my prof that the best way to find the determinant of a matrix is to row reduce it to upper triangular and then take the product of the numbers on the diagonal. That's fine, BUT, how do you know how to reduce it?

Depending on what row operations you do, you get different numbers on the diagonal, and a different product..

How do I go about this?

 

[Office_Shredder]

You should never get a different product, because the determinant doesn't change when you do row operations. If you type up the two different ways that you row reduced it, we can tell you where you made your mistake

 

[theRukus]

Perhaps I'm understanding how to reduce to upper triangular form wrong.

This first one comes from my Professor, so I'm assuming it is right:

0 2 1
-3 2 1
3 6 9

3 6 9 (rearrange rows)
-3 2 1
0 2 1

1 2 3 (r1 / 3)
-3 2 1
0 2 1

1 2 3
0 8 10 (r2 + 3(r1))
0 2 1

1 2 3
0 2 1
0 8 10 (swap r2 and r3)

1 2 3
0 2 1
0 0 6 (r3 - 4(r2))

Okay, so that's alright then. The determinant of this matrix is (6)(2)(1) = 12.

What if I backtrack two steps, and don't swap r2 and r3?

1 2 3
0 8 10
0 2 1

1 2 3
0 8 10
0 0 -6 (4(r3) - r2)

So now I'm stuck. I guess this isn't reduced enough to determine the determinant. Where do I go from here? In general, what is required of an upper triangular matrix in order to find the determinant? I don't understand why this one is not acceptable, and I guess that's where I'm going wrong.

Thank you so much.

 

[Office_Shredder]

Replacing r(3) with 4r(3)-r(2) isn't a row operation.

Also I lied in my previous post. Row operations do change the determinant, either by a -1 or by multiplying by a scalar if you scale a row. Sorry about that

 

[mathwonk]

multiplying a row by a non zero scalar multiplies the determinant by that same scalar. interchanging two rows multiplies the determinant by -1. Adding a scalar multiple of one row to a different row does not change the determinant. With these rules you can somewhat tediously keep track of how the determinant changes under row reduction.

E.g. Replacing r(3) with 4r(3)-r(2) is a combination of two row operations that multiplies the determinant by 4.

If all you want to know however is whether the matrix is invertible, you don't need to keep track. it will just depend on whether any zeroes wind up on the diagonal at the end of the reduction process. E.g. if you ever get a zero row or two dependent rows, the determinant is zero, so stop.

 

[Edgardo]

You have to know all the effects of row operations. Some row operations alter the determinant, e.g. multiplying a row by a scalar c also multiplies the determinant by c.

To keep track of these changes you can write the inverse of the scalar in front of the determinant, e.g.
det(A) = 1/c *det(B) where B has one row multiplied by c in comparison to A.

Example:

<br /> \begin{vmatrix}<br /> 4 &amp; 1 \\<br /> 3 &amp; 5 \\<br /> \end{vmatrix} = <br /> <br /> \frac{1}{2}<br /> <br /> \begin{vmatrix}<br /> 8 &amp; 2 \\<br /> 3 &amp; 5 \\<br /> \end{vmatrix} <br /> <br />

Here, I have multiplied the first row by c=2, thus I have to multiply the determinant by 1/c=1/2.

This is explained very detailed in Paul's Online Math Notes, section http://tutorial.math.lamar.edu/Classes/LinAlg/DeterminantByRowReduction.aspx" .


More material:

1) http://ltcconline.net/greenl/courses/103b/matrices/DETINV.HTM"

2) http://www.youtube.com/watch?v=iOzVZCLioZI"
A youtube video by burny1 demonstrating all rules by an example.

3) http://www.youtube.com/watch?v=32rdijPB-rA"
A youtube video by Salman Khan demonstrating mulitplication of one row for a 3x3 determinant and the general case.

 

[theRukus]

Thanks heaps for all the help guys, that got my issue cleared up!