Calculating Determinant (of a square matrix).

**Timbuk2** · 13-01-2011 11:55PM #1

I've slipped up somewhere here.

You'll have to forgive me - I'm not yet able to do matrices using LaTeX so I've just scanned what I did this morning - I hope that isn't a problem.

Firstly, I calculated the determinant the 'usual' way. Note that the | | signs indicate the determinant of the matrix inside them.

I then did it a second way, by reducing the matrix to upper triangular form. Elementary row operations affect the determinant as follows
-Swapping two rows changes the sign of the determinant
-Adding a scalar multiple of one row to another leaves the determinant unchanged
-If a row is multiplied by a scalar k, the determinant is also multiplied by k

Note the square brackets just denote what E.R.O's I've performed.

Why am I not getting the same answers? The first answer is correct, as far as I know, but the second one is not. I could have stopped where the ** is, and got the determinant as -3, but by further simplifying, I get a different answer. I think my problem is multiplying by the -1/3.

Thanks

Michael Collins · 14-01-2011 12:09AM

You multiplied the last row by -1/3, so to compensate you must multiply the determinant by -3, you have the right idea, just the wrong way around.

**Timbuk2** · 14-01-2011 12:11AM

Michael Collins wrote: »

You multiplied the last row by -1/3, so to compensate you must multiply the determinant by -3, you have the right idea, just the wrong way around.

Oh

That makes complete sense! Thanks!

seandoiler · 14-01-2011 12:13AM

matrix written by (**) is already in upper triangular form, so the determinant would be -3 as you calculated (the further reduction you make works but you have multiplied it out wrong...it should be that the determinant of that matrix is (-1/3)*determinant of the original matrix and not the other way around as you've written it...)

i.e. (-1/3)*det(A)=det(B), where B is that final matrix with the ones along the diagonal

EDIT: Me and my slow typing!!

Michael Collins · 14-01-2011 12:16AM

No problem.

Just to clarify, the statement "if a single row is multiplyed by the scalar k, then so is the determinant", this is meant as a consequence of your action, not a remedy!

So if A is the original matrx, the determinant of your final matrix above will be k . det(A) = -1/3 * det(A).

Since it's actually det(A) you want, you need to divide by k i.e. multiply by 1/k = -3 in this case.

Edit: Damn, seandoiler gets his own back!

**Timbuk2** · 14-01-2011 12:21AM

Thanks! I see where I went wrong now.

I know the further reduction past (**) wasn't necessary, but I did it to include the scalar multiplication, as that what was causing my confusion. Sometimes you have to use scalar multiplication to get it to upper triangular form, it just so happened that this example didn't need it!

The notation that I was using is a bit sloppy and probably is more confusing than clear. I should really call the above det(B) and then at the end, relate it to det(A).

Thanks again!

Calculating Determinant (of a square matrix).

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