# Help with finding a determinant's general form

1. May 28, 2014

### BiGyElLoWhAt

For a homework assignment I need to find the determinant of:
$\left | \begin{array}{cccc} a^0 & b^0 & c^0 & d^0 \\ a^1 & b^1 & c^1 & d^1 \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \\ \end{array} \right |$

I've reduced my matrix, through linear combinations and column operations, 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. I'm so close, but apparently I no kan algebreh.

2. May 28, 2014

### BiGyElLoWhAt

LOL
or should I just multiply the 3rd row by $\frac{(c^2 +ca+a^2 )−(b^2 +ba+a^2 )}{(c+a)−(b+a)}$ and subtract if from the 3rd row and say here you go prof, enjoi.

3. May 28, 2014

### verty

Use the Laplace expansion. Always use the Laplace expansion unless you are told not to, it is, I think, the best method to use.

4. May 28, 2014

### Fredrik

Staff Emeritus
I was thinking that too, although I didn't know it was called Laplace expansion until I looked it up on Wikipedia a few minutes before I saw verty's post. (I've been calling it cofactor expansion). This approach looks promising here.

Does the problem tell you that a,b,c,d are all non-zero? Does it tell you what to do if one of them is zero? Are you supposed to use the definition $0^0=1$ or just set the entire column to all zeroes?

5. May 28, 2014

### jbunniii

I haven't checked your work so far. But assuming it's correct, why not simplify some of your elements before continuing? For example, $(c+a) - (b+a) = c-b$, and
$$(c^2 + ca + a^2) - (b^2 + ba + a^2) = (c^2 - b^2) + (ca - ba) = (c-b)(c+b) + (c-b)a = (c-b)(a+b+c)$$
Hint for this particular type of matrix: wherever possible, try to reduce to expressions of the form $(x-y)$.

6. May 28, 2014

### BiGyElLoWhAt

aha! jbunniii you've done it again. I tried simplifying the c^2 + ca.... blah blah blah expression, but it looked even less promising. I never even tried to simplify a_3,2... I thought I would be better off leaving it alone, at least as far as finding a common factor was concerned. Thanks everyone

7. May 28, 2014

### BiGyElLoWhAt

I'm assuming we're not supposed to, as this was the new tequnique that we learned, and this is the method he used to solve the 3x3 version, and we're supposed to solve the 4x4. otherwise i would've cancelled out the first row and called it a day.

8. May 28, 2014

### Fredrik

Staff Emeritus
If he has explained the method, then you should be allowed to use it. Yes, expand along the first row is what I had in mind. Then you can use the same trick on the 3x3 matrices you end up with.

9. May 28, 2014

### Ray Vickson

Have you looked at LcKurtz's very clever solution of this problem over in the Precalculus forum?

10. May 28, 2014

### LCKurtz

So now we find that this was a duplicate posting by BigYellowHat. Very annoying and violates forum rules. The original thread was:

11. May 29, 2014

### BiGyElLoWhAt

Sorry guys, I posted it there, but then I thought it was better suited here. For future reference is there a way I can move a thread rather than reposting?

12. May 29, 2014

### Fredrik

Staff Emeritus
Use the report button to let the mentors know that you think it should be moved. The report you submit will be the first post of a new thread in the (hidden) mentors' forum. They will do what you ask if they agree that it's appropriate.

13. May 29, 2014

### BiGyElLoWhAt

Thanks Fredrik, so is it ok if I merge the last post from that thread with this thread then? I would like to continue the discussion that was happening there.

14. May 29, 2014

### BiGyElLoWhAt

Here's what I was wondering.

15. May 29, 2014

### jbunniii

It's a good observation, and it gives you a strong hint about what the full expression for the determinant looks like. However, to find the answer definitively, it is sufficient to view just one of the columns as powers of a variable. As LCKurtz's excellent hint showed, if
$$P(x)=\left | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a & b & c & x \\ a^2 & b^2 & c^2 & x^2 \\ a^3 & b^3 & c^3 & x^3 \\ \end{array} \right |$$
then $P(x)$ must be of the form $A(x-a)(x-b)(x-c)$. You should be able to work out what $A$ must be directly from this, without changing which column you treat as variable. Hint: if you compute the cofactor expansion, what is the coefficient of $x^3$?

16. May 29, 2014

### BiGyElLoWhAt

it looks like (bc^2 - cb^2) - (ac^2 - ca^2) + (ab^2 -ba^2).

That's just from the minors that give me x^3 terms, starting from the matrix above. If I expanded the plynomial above I would get Ax^3, but then by cofactor expansion (assuming I'm interpreting this correctly), A would have to be equal to those terms above, but I'm pretty sure that's not correct.

Let me walk you through what I'm thinking, because I'm just not seeing it, and I have no idea why.
To answer your hint question Jbunniii, since the first row is all 1's then my first cofactor expansion will give me 4 3x3 matrices each with a multiple of 1. Since, to answer your question, I'm looking for the coefficient of x^3, I can ignore the abc a^2b^2c^2... matrix because that doesn't contribute any x terms, cubed or otherwise.

This leaves me 3: 3x3; the bcx, the acx, and the abx matrices. performing my last expansion on this I end up with a matrix multiple b |c^2 x^2...| which gives a bc^2 x^3 term. The matrix is positive (from the first expansion) and the b term is positive (second expansion). Then I have a c |b^2 x^2...| which gives me a cb^2 x^3 term, but the c is negative by the (-1)^{i+j} part of the cofactor, and I can ignore the x|b^2 c^2...| minor because it doesn't contribute an x^3 term. This leaves me with (bc^2 - cb^2) from the first minor in my second expansion. The other terms follow by the same logic.

So right now, as it stands, x^3 has a coefficient of A. Through cofactor expansion I find that the coefficient should be c^2(b-a) + b^2(a-c) + a^2(c-b).
This seems excessive, but somewhat reasonable (if that makes any sense), as, in order to satisfy the equality I mentioned earlier, we would need something mildly excessive; but I don't see how I was supposed to see that from what we had without doing the cofactor expansion.

...Unless the substitution was only intended to shorten up the process, NOT to bypass the alternative methods?

17. May 29, 2014

### LCKurtz

If you leave the 3x3 cofactor that is the coefficient of $x^3$ unexpanded, do you see that it is the same kind of problem the 4x4 matrix is? Use the same idea on it.

18. May 29, 2014

### jbunniii

You can use any row or column of the matrix to perform the cofactor expansion. What happens if you use the right column instead of the top row? It should make it possible to see the coefficient of $x^3$ at a glance.

19. May 29, 2014

### BiGyElLoWhAt

Ok so let me see if I'm getting this correctly:

We started this with :
$M= \left | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \\ \end{array} \right |$
and we decided to solve a similar problem, instead, by defining :
$P(x) =\ \text{det} \left | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a & b & c & x \\ a^2 & b^2 & c^2 & x^2 \\ a^3 & b^3 & c^3 & x^3 \\ \end{array} \right |$
and knowing that the solution to our problem was the solution to $P(d)$.
We know $P(a)=P(b)=P(c)=0$.
So this leaves $A(x-a)(x-b)(x-c) = P(x)$;
I don't quite understand where the A came from, other than the fact that without the A, it's not dimentionally correct. Using a vector interpretation of our matrix, a volume in $\mathbb{R}^4$ should have units4, is this part of that factor theorem you were talking about? I don't think I've heard of it, I'm in my second week of a 6 week summer linear algebra course that crams a whole semester in that time. I have to go for now, I'll be back on in a short while. Thanks for the help everyone.

* * * * * * *

[part 2]
Ok so if we have $\text{det} P(x) =\ \text{det}M = A(x-a)(x-b)(x-c) \ \text{&} \ P(x) = \text{solve through cofactor expansion}$
then we know by expanding our factored polynomial that we will end up with a term $Ax^3$.
So by solving for the coefficient of $x^3$ through cofactor expansion, which, as LC pointed, can be expressed as such:

$\text{det} \ \left | \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \\ \end{array} \right | x^3$
Now, add another iteration of this method:
Define:
$Q(y) = \text{det} \ \left | \begin{array}{ccc} 1 & 1 & 1 \\ a & b & y \\ a^2 & b^2 & y^2 \\ \end{array} \right |$
(knowing that $Q(c)$ is what we're looking for)
We are left with $\text{det} \ M = P(d) = A (d-a)(d-b)(d-c) = Q(c)(d-a)(d-b)(d-c)$

So from $Q(y)$ we can gather that $Q(a)=Q(b)=0$ therefore we can say that
$Q(y) = B(y-a)(y-b)$
Looking good so far?
...
Now by the same logic, we can solve for B:
$B = \text{det} \ \left | \begin{array}{cc} 1 & 1 \\ a & b \\ \end{array} \right |$
Which is highly solvable.

So in the end, we have
$B = (b-a)$
$Q(c) = (b-a)(c-a)(c-b)$
$P(d) = (b-a)(c-a)(c-b)(d-a)(d-b)(d-c) = \text{det} \ M$

How's that? =]

The only thing is, and maybe this stems from the fact that this is just a calculation and not a geometry problem, this is not what I would call dimentionally sound. If the determinant of a matrix is the n-space volume of the nth dimentional vectors that comprise it, how come I have 4d vectors and a 6th dimentional volume? Wiiieeeerrdd, but I think I solved it properly.

Last edited: May 29, 2014
20. May 29, 2014

### jbunniii

It comes from these two facts:
• P(x) is a cubic polynomial
• P(x) has roots $a$,$b$, and $c$
Therefore $P(x)$ must be of the form $A(x-a)(x-b)(x-c)$. We need the $A$ because any polynomial of the form $A(x-a)(x-b)(x-c)$ satisfies both conditions. So we need to extract more information from the matrix in order to determine the value of $A$.