# I need a multiple for a linear combination

1. May 28, 2014

### BiGyElLoWhAt

I'm working with matrices, so maybe this should be in the calc and beyond, but my problem is with the algebra (-.-)

I've reduced my matrix to this:

$(b-a)(c-a)\left | \begin{array}{cccc} 1 & (b-a)^{-1} & (c-a)^{-1} & 1 \\ 0 & 1 & 1 & (d-a) \\ 0 & 0 & (c+a) - (b+a) & (d^2- a^2) - (b + a)(d - a) \\ 0 & 0 & (c^2 + ca +a^2) - (b^2 +ba +a^2) & (d^3 - a^3)- (b^2 + ba + a^2)(d - a) \\ \end{array} \right |$

Now here's my dilemma, I need $a_{4\ 3} = 0$, which means that I need to multiply $[(c+a) - (b+a)]$ by something to get that ugly term $(c^2 + ca +a^2) - (b^2 +ba +a^2)$...

I'm not seeing anything, but I'm sure it can be done.

2. May 28, 2014

### LCKurtz

What did your determinant look like when you started?

3. May 28, 2014

### BiGyElLoWhAt

$\left | \begin{array}{cccc} a^0 & b^0 & c^0 & d^0 \\ a^1 & b^1 & c^1 & d^1 \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \\ \end{array} \right |$

Last edited: May 28, 2014
4. May 28, 2014

### LCKurtz

I will show you a trick that you likely wouldn't think of. Let's write that as
Now let's replace the $d$'s by $x$'s and call it $P(x)$:$$P(x)=\left | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a & b & c & x \\ a^2 & b^2 & c^2 & x^2 \\ a^3 & b^3 & c^3 & x^3 \\ \end{array} \right |$$
Do you see that $P(x)$ is a cubic polynomial? Then do you see that $P(a)=P(b)=P(c)=0$? So by the factor theorem $P(x) = A(x-a)(x-b)(x-c)$? Do you see what $A$ would have to be? All these questions can be answered by looking at the determinant without expanding it.

That might give you some ideas. I have to leave for now though. That determinant is called a Vandermonde determinant and you will find more than you want to know if you Google it.

[Edit, added later in the day]Sorry I had to leave but I will be in and out this evening. I did want to add that if you put $x=d$ back in your original determinant you now have$$P(d) = \left | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \\ \end{array} \right | = A(d-a)(d-b)(d-c)$$and $A$ is the coefficient of $x^3$ in $P(x)$. Are you with me so far?

Last edited: May 28, 2014
5. May 28, 2014

### willem2

you can simplify both (c+a) - (b+a) and (c^2+ca+a^2)−(b^2+ba+a^2) and then it will be easy to see that the first term divides the second.

6. May 29, 2014

### BiGyElLoWhAt

Well, assuming we're defining P(x) to be the determinant of the matrix, then yes, I see that P(a)=P(b)=P(c)=0, by 1) the scalar multiple theorem for determinants, or whatever its called, and also, as of recently I've been enlightened as broseph the geometric interpretation of a determinant, which I guess is somewhat causal of the scalar multiple theorem.

As far as the factor theorem goes, I see what you're doing as far as factoring out the 0's of the polynomial; that's just basic algebra. I'm not sure what this A is that you're getting at. Maybe I should spend some time on Google.

But... with this method, couldn't we do the same with the A's and the B's and the C's? My zeros would differ in each case. Is A supposed to account for the equality of the polynomial? So maybe something to the order of

A(d-a)(d-b)(d-c) = B(c-a)(c-b)(c-d) =C(b-a)(b-c)(b-d) ...
or is that not what you're getting at? I guess that's more or less just an observation.

7. May 29, 2014

### Staff: Mentor

That's exactly the reason why you should not open multiple threads for the same topic.