Abstact Matrix Help! No numbers in matrix :(

Printable View

February 20th 2011, 04:28 PM
fizzle45

Abstact Matrix Help! No numbers in matrix :(

$\text{if } a\neq b\neq c\text{ and the equations}$
$ax+a^2y+(a^3+1)=0$
$bx+b^2y+(b^3+1)=0$
$cx+c^2y+(c^3+1)=0$

$\text{are consistent, then prove that }abc=-1.$

I could use some desperately needed help on this one. My first plan of attack was to move all the terms with no variables onto one side. This gives us a system of equations with 3 equations but with only two unknowns,(I think the key is in this detail?) Does this mean that if I take any two equations, the one with a's and b's for example, and find the x and y solution that those two solutions will have to equal another x y solution I obtained from maybe using the equations with just b's and c's?
I have a feeling I'm totally off base. No one in my class has gotten this problem so anything would be helpful thanks.
February 21st 2011, 02:55 AM
Ackbeet

Well, this is a long and complex proof, but it gets the job done, I think.

1. Consistency, in your case, means that

$\left|\begin{matrix}a&a^{2}\\ b&b^{2}\end{matrix}\right|=0$ implies $\left|\begin{matrix}a^{2} &-a^{3}-1\\ b^{2} &-b^{3}-1\end{matrix}\right|=0.$ For a justification of this statement, see this thread - all of it.

2. Computing these determinants means you get that

$ab^{2}-a^{2}b=0\quad\Rightarrow\quad -a^{2}b^{3}-a^{2}+b^{2}a^{3}+b^{2}=0.$

3. However, $ab^{2}-a^{2}b=0$ is entirely equivalent to $ab(b-a)=0,$ which is entirely equivalent to $ab=0,$ since $b-a\not=0$ by assumption. Hence, if the "if" part is true, then either a or b is zero.

4. But if a or b is zero, the implication means that the other one of a or b is also zero, in contradiction with $a\not=b.$ Hence, $ab\not=0.$

5. Thus, the determinant of the usual coefficient matrix associated with the first two equations is nonzero, hence the matrix will have a unique solution. Thus, you should, at this point, solve the system.

6. Now simplify the expression

$cx+c^{2}y+c^{3}+1=0$ with your plugged-in values for x and y.

I get

$(a-c)(b-c)(1+abc)=0.$

The conclusion follows.

Does that make sense?
February 21st 2011, 06:34 AM
fizzle45

Det question

$\text{if } a\neq b\neq c\text{ and the equations}$
$ax+a^2y+(a^3+1)=0$
$bx+b^2y+(b^3+1)=0$
$cx+c^2y+(c^3+1)=0$

$\text{are consistent, then why }\left|\begin{array}{ccr}a&a^2&(a^3+1)\\b&b^2&(b^3 +1)\\c&c^2&(c^3+1)\end{array}\right|=0$
February 21st 2011, 06:40 AM
FernandoRevilla

The rank of the matrix associated to the unkmowns is at most $2$ so, ...

Fernando Revilla
February 21st 2011, 06:42 AM
Ackbeet

Actually, I'm not sure that that particular determinant is zero. What I did say, was that if you have the consistent system

$ax+a^{2}y=-a^{3}-1$
$bx+b^{2}y=-b^{3}-1,$

then

$\left|\begin{matrix}a&a^{2}\\b&b^{2}\end{matrix}\r ight|=0$ implies that it must be the case that

$\left|\begin{matrix}a^{2}&-a^{3}-1\\b^{2}&-b^{3}-1\end{matrix}\right|=0.$

The reference for this fact is in post # 2.

Does that make sense?
February 21st 2011, 06:55 AM
fizzle45

no :( i feel so stupid right now. the solution i recieved looks nice and pretty but it uses the fact that the 3x3 system i posted has det 0 but doesn't say why, so im stuck on that. what about $\left|\begin{array}{ccr}a&a^2&-a^3-1\\b&b^2&-b^3-1\\c&c^2&-c^3-1\end{array}\right|$ can we say this equals 0?
February 21st 2011, 07:02 AM
Ackbeet

In revisiting Fernando's Post # 4, I can see why the determinant you produce there must be zero. For an overdetermined system to be consistent, at least two rows must be linearly dependent, right? It just so happens that in your system, you only have one extra equation. That makes it possible to form the determinant you wrote down. Two rows of that matrix must be linearly dependent, or you'd have an inconsistent system. Therefore, that determinant is zero.