Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$

Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.

My textbook only provides two simple examples.

Really have no idea how to do this type of questions..

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I tried to take row 1 from row2 and row3, so I get a 2x2 determinant. {{b^3+b^2-a^3-a^2,b-a},{c^3+c^2-a^3-a^2,c-a}}.Then I tried to factorize b^3+b^2-a^3-a^2 to get (b-a)*(something). – Vic. Jul 21 '12 at 10:40
I don't know how to write determinant in LaTeX so I didn't write what I've tried in the question cause I think it might be difficult to read. – Vic. Jul 21 '12 at 10:44
I will not write the answer I prepared cause I think it might be difficult to read. – Did Jul 21 '12 at 10:50
Me? I think nothing, I just noted that what you wrote (1.) is absurd and (2.) goes against explicit recommendations about how to ask questions on this site. – Did Jul 21 '12 at 16:56
You could both of you afford to be a little less... accusatory? I agree with did inasmuch as if you wrote something ugly, Vic, someone would just come along and fix the formatting anyway. – Ben Millwood Jul 22 '12 at 23:25

3 Answers

$\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$

$=\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2-(a^3+a^2) & b-a & 1-1 \\ c^3+c^2-(a^3+a^2) & c-a &1-1\end{pmatrix}$ (applying $R_2'=R_2-R_1\ and\ R_3'=R_3-R_1$)

$=(b-a)(c-a) \det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^2+a^2+ab+b+a & 1 & 0 \\ c^2+a^2+ca+c+a & 1 & 0\end{pmatrix}$

$=(b-a)(c-a)\cdot1\cdot \det\begin{pmatrix} b^2+a^2+ab+b+a & 1 & \\ c^2+a^2+ca+c+a & 1\end{pmatrix}$

$=(b-a)(c-a)(b-c)(a+b+c+1)$

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Thanks for answering, I know how to get to step 2, but failed to factorize (b-a) out of (b^3+b^2-a^3-a^2). I'm going to do some exercise of division of polynomial. – Vic. Jul 21 '12 at 11:20
@Vic: Exercises on doing are good! Here's a tip on recognition: by using general knowledge of polynomials, $(b-a)$ divides a polynomial in $a$ and $b$ if you get zero by replacing $b$ with $a$. A common way this happens is when you have an expression with the same polynomial, but once in $a$ and once in $b$. That is, $(b-a)$ divides $f(b) - f(a)$ whenever $f$ is a polynomial. Recognizing this can help with computation too, since it suggests grouping terms of like total degree -- e.g. factor $(b-a)$ out of $b^3 - a^3$ and out of $b^2 - a^2$ separately, then combine the results. – Hurkyl Jul 21 '12 at 17:17
The $R_1\ and\ R_2$ becomes identical if a is replaced with b or vice versa. So, (a-b) is a factor of the given determinant. So are (b-c),(c-a). Now, the given determinant is symmetric in (x,y,z) and of degree 4. So, the value should be of the form (a-b)(b-c)(c-a)(p(a+b+c)+q) where p,q are constants independent of a,b,c. Comparing the co-efficient of $a^3$, we get (b-c)=(b-c)(-1)p=>p=-1 Comparing the co-efficient of $a^2$, we get (b-c)=(b-c)(-1)q=>q=-1 So, the given determinant = -(a-b)(b-c)(c-a)((a+b+c)+1) – lab bhattacharjee Jul 22 '12 at 15:59
That's pretty much exactly the approach my answer below took. – Ben Millwood Jul 22 '12 at 23:26
@BenMillwood, the expression is not homogeneous unless we break into the 2 parts as pre-kidney did below, right? – lab bhattacharjee Jul 23 '12 at 4:34

Regard $b$ and $c$ as constants, and the determinant as a polynomial in $a$. Then find ways of making the determinant equal to 0, and by the Factor Theorem you'll get a factor of the determinant.

Obvious choices: set $a=b$ or $a=c$ and you'll have two identical rows, so $(a-b)$ and $(a-c)$ are factors.

We can clearly see that permuting the variables is the same as permuting the rows, and hence only changes the determinant by a sign. Hence permuting the variables in any factor gives another factor, so $(b-c)$ is a factor.

What's left? Well, we can see without much difficulty that the determinant is cubic in $a$, and the coefficient of $a^3$ is $(b-c)$. We've already taken out a factor of $(b-c)$ and two monic linear factors, so whatever's left is linear and monic in $a$ and, by similar arguments, must also be so in $b$ and $c$. Hence it pretty much has to be $a + b + c + k$ for some constant $k$. So we've got as far as the following: $$\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}=(a-b)(a-c)(b-c)(a+b+c+k)$$ for some $k$, and all we need to do is find $k$. I first tried setting $b=c=0$ but that just leaves you with $0k = 0$, so instead let's set $b=1$, $c=0$: \begin{align} \det\begin{pmatrix} a^3+a^2 & a & 1 \\ 2 & 1 & 1 \\ 0 & 0 &1\end{pmatrix}&=a(a-1)(a+1+k) \\ \det\begin{pmatrix} a^3+a^2 & a \\ 2 & 1\end{pmatrix}&=a(a-1)(a+1+k) \\ a^3 + a^2 - 2a &=a(a-1)(a+1+k) \\ a^2 + a - 2 &= (a-1)(a+1+k)\end{align} Comparing constant terms, $-2 = -1-k$ so $k=1$, and we're done!

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\begin{align*} \det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix} & = \det\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c &1\end{pmatrix}+\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}\\ & = -(a-b)(b-c)(c-a) -(a-b)(b-c)(c-a)(a+b+c)\\ &= (a-b)(b-c)(a-c)(a+b+c+1) \end{align*}

Hey guys, editing my post to address concerns. You're right, I didn't put in all the details because I thought the method was clear once you've seen Vandermonde Determinants. Here it is explicitly: \begin{align*} \det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix} & = \det\begin{pmatrix} a^3+(b+c)a^2 & a & 1 \\ b^3 +(a+c)b^2& b & 1 \\ c^3+(a+b)c^2 & c &1\end{pmatrix}\\ & = \sum_{cyc}\det\begin{pmatrix} a^3 & a & 1 \\ ab^2 & b & 1 \\ ac^2 & c &1\end{pmatrix} & \\ &= \sum_{cyc}a\det\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c &1\end{pmatrix} \end{align*} (At this point we have shown the result.)

Okay, so I guessed intuitively that $$\det\begin{pmatrix} (b+c)a^2 & a & 1 \\ (a+c)b^2& b & 1 \\ (a+b)c^2 & c &1\end{pmatrix}=0$$But its easy to prove, by finding the eigenvector: $$\begin{pmatrix}1 \\ -(ab+bc+ca) \\ abc\end{pmatrix}$$

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Despite the 4 upvotes, the page you link to DOES NOT allow to compute the second determinant on the RHS, hence this step of the proof is without justification. You might want to explain how you know this value. – Did Jul 22 '12 at 7:33
Explained; see post. – pre-kidney Jul 22 '12 at 20:28