Determinant with Very Hard Algebra

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August 7th 2007, 01:36 AM
binarybob0001

Determinant with Very Hard Algebra

The problem is
$ \det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ] $
Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.
$ abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b) $
There are many ways to factor out terms from this.
$ abc[bc(c - b) + ac(a - c) + ab(b - a)] = $
$ abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] = $
$ abc[c^2(b - a) + b^2(a - c) + a^2(c - b)] $
There are even more ways of course. None of these ways lead me to the simple solution:
$ abc(b - a)(c - a)(c - b) $
Can someone explain how to factor this? Thanks.
August 7th 2007, 02:43 AM
JakeD

Quote:

Originally Posted by binarybob0001

The problem is
$ \det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ] $
Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.
$ abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b) $
There are many ways to factor out terms from this.
$ abc[bc(c - b) + ac(a - c) + ab(b - a)] = $
$ abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] = $
$ abc[c^2(b - a) + b^2(a - c) + a^2(c - b)] $
There are even more ways of course. None of these ways lead me to the simple solution:
$ abc(b - a)(c - a)(c - b) $
Can someone explain how to factor this? Thanks.

Multiply out

$\begin{aligned}(b - a)(c - a)(c - b) &= (bc -ac -ab +a^2)(c-b) \\ &= bc^2 - ac^2 -abc + a^2c -b^2c +abc + ab^2 - a^2b \\ &= bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b \end{aligned} $

as you wanted.
August 7th 2007, 06:28 AM
binarybob0001

You can only do that if you already know the solution. What if you didn't know the solution? I would have just stopped thinking I was done. Is there a systematic way of arriving at that result?
August 7th 2007, 08:06 AM
Soroban

Hello, binarybob0001!

Quote:

The problem is: . $ \begin{bmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{bmatrix}$

Sorry I'm new. . . . . Welcome aboard!
That's the best I could write the problem. . . . . Good job!

It's easy to find the determinant with expansion by minors
but to get the simplified solution involves some sort of clever trick that I don't know of.

Here's what I can get to:
. . $abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)$

There are many ways to factor out terms from this.
None of these ways lead me to the simple solution:
. . $abc(b - a)(c - a)(c - b)$
Can someone explain how to factor this?

Here's an approach I "discovered" years ago . . .

Collect terms with respect to one of the variables.

Let's use $c$
. . Collect terms with $c^2$ , and with $c$ , and 'constants' (no $c$ 's).

We have: . $abc\bigg[bc^2 - ac^2 - b^2c + a^2c + ab^2 - a^2b\bigg]$

Factor "by grouping": . $abc\bigg[c^2(b-a) - c(b^2-a^2) + ab(b-a)\bigg]$
. . $= \;abc\bigg[c^2(b-a) - c(b-a)(b+a) + ab(b-a)\bigg]$

Factor out $(b-a)\!:\;\;abc(b-a)\bigg[c^2 - c(b+a) + ab\bigg]$
. . $= \;abc(b-a)\bigg[c^2 - bc - ac + ab\bigg]$

Factor "by grouping": . $abc(b-a)\bigg[c(c-b) - a(c-b)\bigg]$

Factor out $(c-b)\!:\;\;abc(b-a)(c-b)(c-a)$ . . . . ta-DAA!
August 8th 2007, 12:35 AM
binarybob0001

ahah, so I just need to chose a letter, stupid me. Thank you very much. Hey, are there more problems this hard to practice on?