Math Is Fun Forum

kendricktamis

Member

Registered: 2010-04-06

Posts: 4

Trying to solve a linear algebra matrix problem?

Ok, so I am trying to prove that one matrix equals another without finding determinates. The first matrix is
(1, a, bc)
(1, b, ac)
(1, c, ab)

I need to figure out how to make that equal
(1, a, a^2)
(1, b, b^2)
(1, c, c^2)

I need to do this using column operations, so far I've tried multiplying the 2nd column by abc and subtracting the 3rd column to get the 3rd column, but I can't figure out how to finish off and get the squared letters by themselves. Thanks a ton for anyone willing to try to figure this thing out.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Trying to solve a linear algebra matrix problem?

Hi kendricktamis;

I really don't see how you can answer that by column operations.

For two matrices to be equal they must be the same size and every element of one matrix must me the same as every element of the other. Since the first 2 columns of both matrices are the same you are left with solving a 3 x 3 non linear simultaneous set.

You must solve for a,b,c in the system:

b*c = a^2
a*c = b^2
a*b = c^2

This is not easy, and is best done with a Computer Algebra System. A trivial answer is a=b=c.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Trying to solve a linear algebra matrix problem?

b = a^2/c, so a*c = a^4/c^2 or rather a^3 = c^3 meaning a = c.  Thus b = a.  The only solution in this case is the trivial one.  The other case is where my previous work does not apply, c = 0 (or a = 0, but by symmetry, you need to only consider one case).  But if this were the case, then bc = a^2 and ac = b^2 assure that both a and b are zero as well.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."