Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

there's this determinant problem I've been working on for several days now whose answer I can't quite get to:

$$ D = \left| \begin{array}{ccc} a^3+a^2 & a & 1\\ b^3+b^2 & b & 1\\ c^3+c^2 & c & 1\\ \end{array} \right| $$

Express the determinant as the product of four linear factors.

The given answer is $(a-b)(b-c)(c-a)(a+b+c+1)$ but I'm stuck after getting the first two factors, $(a-b)$ and $(b-c)$:

$$ D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & a-b & 0\\ b^2+bc+c^2 & b-c & 0\\ c^3+c^2 & c & 1\\ \end{array} \right| $$

whose determinant transposes into:

$$ D=(a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & b^2+bc+c^2 & c^3+c^2\\ 1 & 1 & c\\ 0 & 0 & 1\\ \end {array} \right| \\ \Rightarrow D=(a-b)(b-c))1(a^2+ab+b^2)-1(b^2+bc+c^2))\\ =(a-b)(b-c)(a^2+ab+bc+c^2) $$

This is where I am stuck.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Playing around, and beginning pretty much as you began (except carrying out the row operations correctly): $$ D = \begin{vmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b & 0 \\ (b^3-c^3)+(b^2-c^2) & b-c & 0 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b \\ (b^3-c^3)+(b^2-c^2) & b-c \end{vmatrix}\\ = (a-b)(b-c) \begin{vmatrix} a^2+ab+b^2+a+b & 1 \\ b^2+bc+c^2+b+c & 1 \end{vmatrix}\\ = (a-b)(b-c)(a^2+ab+b^2+a+b-b^2-bc-c^2-b-c)\\ = (a-b)(b-c)(a^2+ab+a-bc-c^2-c)\\ = (a-b)(b-c)(a^2+ab+a+ac-ac-bc-c^2-c)\\ = (a-b)(b-c)(a-c)(a+b+c+1), $$ which, annoyingly enough, is minus the answer you're given.

share|improve this answer
    
Indeed, thank you! –  Branimir Ćaćić Mar 18 '13 at 8:52

Hint: Sorry, you did a mistake at $D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & a-b & 0\\ b^2+bc+c^2 & b-c & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$. It must be $D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & 1 & 0\\ b^2+bc+c^2 & 1 & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$.

Now, what is the determinant if you expand along the third column? It is a $2\times 2$ determinant $\left| \begin{array}{cc} a^2+ab+b^2 & 1 \\ b^2+bc+c^2 & 1 \\ \end{array} \right|$.

Also, though your approach is fine, there may be alternative ways as well.


Edit: Here is some more hints to find the determinant (possibly "without expanding"):

First note that $D=D_1+D_2$, where $D_1 = \left| \begin{array}{ccc} a^3 & a & 1\\ b^3 & b & 1\\ c^3 & c & 1\\ \end{array} \right|$. Now to find the value of $D_1$, note that if we put $a=b$, or $b=c$, or $c=a$, the determinant vanishes. So, $(a-b)(b-c)(c-a)$ must be factors of the determinant. Now, note that the determinant has a leading term (=product of diagonal entries) $a^3b$. So, the other factor must be a linear...and finally the determinant is actually $-(a-b)(b-c)(c-a)(a+b+c)$. Apply the same technique to find $D_2$ and then add.

share|improve this answer
    
In any case, your transposition step is not correct. How did you get it? –  Tapu Mar 18 '13 at 8:06
    
This also gives the same value when expanded, so I'm not too sure where to start again because I do want those $(a-b)$ and $(b-c)$ factors. –  Harris Crescendo Mar 18 '13 at 8:20
    
Please see my Edits...for another way. –  Tapu Mar 18 '13 at 8:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.