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so, I've been reading up on alot of linear algebra lately, and I've been seeing alot of uses for determinants. I learned how to calculate using them and all that good stuff but, I have not been able to figure out what it is exactly, is there some sort of geometric representation of it or anything else like that?

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it represents the proportional 'volume' change. volume depends on the contex. eg in 1 dimension it is length, two dimensions it is area 3 it is volume as we think of it in general, and in higher dimensions it is the analogue of these quantities. if you take the segment of the that is the interval [0,1], or the square with corners (0,0), (1,0) (0,1) and (1,1) and in three d space the cube with volume 1 then the matrix sends this shape to one with volume the determinant of the matrix.

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I see, so the matrix that is formed from those 4 points has the determinant equal to 1, wich is the volume, but then why does it come up so often in linear algebra calculations if its only equal to the area

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that isn't what i said. what matirx formed from those 4 points? i didn't say anything about that. the matrix is a priori given to me (by you).

 

a matrix is invertible if and only if its determinant is noty zero, which is why you need to know it, and to find the inverse (and thus solve the basic linear algebra exercises) it is necessary to know the determinant.

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you mentioned four points and how the area of the square was the going to be 1, it looked like the points could be put into a matrix and it seemed that the matrix (I didn't actually try to find the determinate , just thought that it could be put into reduced row echelon form which would have a determinate of 1) so I thought that you were trying to say something like the determinate could be seen as the volume of the matrix.

 

that last bit is worded horribly, I know but hopefully you can get the gist of what Im saying

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but then why does it come up so often in linear algebra calculations
I'm guessing it comes up very often because a system of equations has non-trivial solutions only if the determinant vanishes. This has great importance in eigenvalue problems.
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Well, geometrically, descriminats can be used by Cramer's Rule which is to find the intersection of 2 lines...where the equations of ad-bc can not =0 or else they are parallel and so forth....(Best I can come up with)

OK, this is what I used to figure it out...

AX+BY=E

CX+DY=F A,B,C,D,E,F are constants not variables

 

Solve for x

ADX+BDY=DE Multiply by D

(-) Bcx+bdy=BF Multiply 2nd equation by B

--------------------

ADX-BCX=DE-BF Subtract

(AD-BC)X=DE-BF Factor

 

X=DE-BF/(AD-BC)

 

Then when you see this, one thing should pop out and that is the denominator, it is the same as the stadard definition of a determinant of AB-BC...

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