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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).2012-07-21
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    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.2012-07-21
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    I will not write the answer I prepared *cause I think it might be difficult to read*.2012-07-21
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    I'm not using this website to finish my homework, I'm self studying further pure math but the textbook only provides simple examples and answers for the exercises. The answer is (a-b)(b-c)(c-a)(a+b+c+1). I'm trying to figure out how to get this. I also tried Wolframalpha(which only gives out the answer) and google for 'determinant factorization' to find if there's a general method/pattern to do this type of question. You think I'm asking this question cause I'm lazy and tried nothing?2012-07-21
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    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.2012-07-21
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    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.2012-07-22

3 Answers 3

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

$=\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.2012-07-21
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    @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.2012-07-21
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    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)2012-07-22
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    That's pretty much exactly the approach my answer below took.2012-07-22
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    @BenMillwood, the expression is not homogeneous unless we break into the 2 parts as pre-kidney did below, right?2012-07-23
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    @labbhattacharjee: indeed it is not. Why is that a problem?2012-07-31
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    @BenMillwood, I'm not sure, we can directly write (a+b+c+k) if it's not homogeneous.2012-08-05
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    @labbhattacharjee: can you provide more detail, please? I see nothing wrong with my reasoning - I'm just comparing coefficients of each power of $a$, and then using symmetry. None of that requires homogeneity.2012-08-05
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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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$$\begin{align*} \det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix} & = \det\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c &1\end{pmatrix}+\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}\\ & = -(a-b)(b-c)(c-a) -(a-b)(b-c)(c-a)(a+b+c)\\ &= (a-b)(b-c)(a-c)(a+b+c+1) \end{align*}$$

see Vandermonde matrix


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.2012-07-22
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    Explained; see post.2012-07-22