# Linear Algebra determinant reduction

Prove, without expanding, that \begin{vmatrix} 1 &a &a^2-bc \\ 1 &b &b^2-ca \\ 1 &c &c^2-ab \end{vmatrix} vanishes.

Any hints ?

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what do you mean with "without expanding"? – Bman72 May 23 '14 at 15:25
not trying to explicitly calculate the determinant (Laplace's expansion) – square_one May 23 '14 at 15:27
You can give a nontrivial linear combination of the columns (or rows) that equals $0$ to show that it vanishes. – Servaes May 23 '14 at 15:29

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

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This is very nice to my honest opinion, the only thing is that the OP is not allowed to use the laplace expansion, but it is enough to leave the first row and column that you correctly canceled because it contained the column $\begin{pmatrix}1 \\0\\0\end{pmatrix}$ – Bman72 May 23 '14 at 15:37
What does "OP" mean... ? I have seen this elsewhere as well ! – square_one May 23 '14 at 15:40
@user148176 the OP is you! the Original Poster – Bman72 May 23 '14 at 15:43
@user148176 Also means 'Original Post' - i.e. the thing or the person behind it, obvious from context. (And it's hard to imagine that the difference even matters, other than for correct grammar). – Ollie Ford May 24 '14 at 2:23
@Ale "We have met the OP and he is us." -Walt Kelly – SpamIAm May 24 '14 at 18:14

Let

$$f(a)=\begin{vmatrix} 1 &a &a^2-bc \\ 1 &b &b^2-ca \\ 1 &c &c^2-ab \end{vmatrix}$$ then it's easy to see that $f$ is a polynomial on $a$ with degree at most $2$ and $f(b)=f(c)=0$ so $$f(a)=\lambda(a-b)(a-c)$$ now, WLOG assume that $bc\ne0$, take $a=0$; we see easily that $f(0)=0$ hence $\lambda=0$.

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Why is $f(0)=0$? – user1551 May 23 '14 at 19:48
Smooth! Nice work. – amWhy May 24 '14 at 11:43

e.g. if your matrix is $A$, consider $(b-c,c-a,a-b) A$

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Add $(ab+bc+ca)$ times the first column to the last and find the common factor of the last column.

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Simply add a multiple of the first column to the last $$\begin{vmatrix} 1 & a & a^2-bc\\1&b&b^2-ac\\1&c&c^2-ab\end{vmatrix} =\begin{vmatrix} 1 & a & a^2-bc+(1)(ab+ac+bc)\\1&b&b^2-ac+(1)(ab+ac+bc)\\1&c&c^2-ab+(1)(ab+ac+bc)\end{vmatrix} = \begin{vmatrix} 1 & a & a(a+b+c)\\1&b&b(a+b+c)\\1&c&c(a+b+c)\end{vmatrix} = 0$$ The final step follows because the second and third column are linearly dependent.

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Just realized this is what LutzL said to do – user153126 May 24 '14 at 17:10