Help with finding a determinant's general form

[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.

 

[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.

 

[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.

 

[Fredrik]

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?

 

[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 $(x-y)$.

 

[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

 

[BiGyElLoWhAt]

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

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 canceled out the first row and called it a day.

 

[Fredrik]

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.

 

[Ray Vickson]

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.

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

 

[LCKurtz]

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

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

https://www.physicsforums.com/showthread.php?t=755660

 

[BiGyElLoWhAt]

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

https://www.physicsforums.com/showthread.php?t=755660

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?

 

[Fredrik]

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?

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.

 

[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.

 

[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.

Here's what I was wondering.

 

[jbunniii]

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.

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 $x^3$?

 

[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?

 

[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.

 

[jbunniii]

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.

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.

 

[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 dimensional vectors that comprise it, how come I have 4d vectors and a 6th dimensional volume? Wiiieeeerrdd, but I think I solved it properly.

 

[jbunniii]

So this leaves $A(x-a)(x-b)(x-c) = P(x)$;
I don't quite understand where the A came from

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 $A$.

 

[BiGyElLoWhAt]

Finished my post above

 

[LCKurtz]

Finished my post above

And you have it solved correctly.

 

[BiGyElLoWhAt]

can you answer my question about the dimensional soundness of the solution with respect to the system? I'm only in my 2nd week of dealing with matrices, and I don't really fully understand what's going on just yet.

 

[jbunniii]

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 dimensional vectors that comprise it, how come I have 4d vectors and a 6th dimensional volume? Wiiieeeerrdd, but I think I solved it properly.

That's a good observation and a good question. I had to think about this for a while.

The answer is that the columns of the matrix contain up to cubic powers of $a,b,c,d$, so $a,b,c,d$ themselves do not represent lengths. Indeed, the determinant represents the (signed) volume of the parallelepiped defined by the column vectors of the matrix. The column vectors are confusing because they mix different powers of $a,b,c,d$. So it's not obvious how the volume should scale with $a,b,c,d$.

Here is a somewhat dubious argument:

If we transpose the matrix, the determinant doesn't change, so an equivalent characterization is that the determinant is the signed volume defined by the row vectors. At least the row vectors have consistent powers of $a,b,c,d$, so if we assume that $a,b,c,d$ all have the same units, then we can understand the units of each row vector.

Now the rows are $(1,1,1,1)$, $(a,b,c,d)$, $(a^2, b^2, c^2, d^2)$, and $(a^3, b^3, c^3, d^3)$. So in the first row, length has same units as $a^0$, in the second row it has same units as $a^1$, and so on. So the volume, which is the product of the four lengths, has the same units as $a^0 \times a^1 \times a^2 \times a^3 = a^{0+1+2+3} = a^6$.

This gives us some insight regarding why you have powers of six appearing in the formula for the determinant.

I repeat, this argument is dubious. The reality is that if $a,b,c,d$ are not unitless, then the four rows do not have the same units, so "volume" is not simply the fourth power of the units of the rows.

 

[BiGyElLoWhAt]

Ahhh, I was wondering about that. So it's not something to the effect of 1 meter x a meters x a^2 meters x a^3 meters.

That's how I was interpreting it, but I like the (a meters)^n better.

 

[BiGyElLoWhAt]

Thanks everyone, and especially thanks to LCKurtz for the clever method of solving this type of matrix. Hopefully I can apply this type of thinking elsewhere in my lifetime of using matrices (and other maths) XD

 

[LCKurtz]

Suppose the $x$ and $y$ axes are measured in feet. You plot the graph of $y=f(x)$ putting $x$ (in feet) in and getting $y$ (in feet) out for the point $(x,y)$ on the graph. Now suppose that $f(x) = 2x + 3x^2 + 5x^3$. You don't think of the units on the right side as feet + feet^2 + feet^3. You do the calculation and express the answer in feet. Otherwise how could any Taylor series be dimensionally sound?

 

[BiGyElLoWhAt]

True. I do find it peculiar that the multiplicitave degree of each column equals the degree of my answer, however.

Anyway, it doesn't really matter, it's just a number problem; there was no meaning behind these numbers other than the fact that they were there to be crunched. I was just wondering what was going on.