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How can I show using properties of determinants that:

$$\det\begin{pmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \\ \end{pmatrix} = 2abc(a+b+c)^3$$

Robert Howard
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JuanMuñoz
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    Starting with Sarrus http://en.wikipedia.org/wiki/Rule_of_Sarrus ?? – MattAllegro Oct 01 '14 at 22:35
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    Sorry i already know the Sarrus law. The matter is to get the result using the properties of determinants like: $det(B)=-det(A)$ if $B$ is $A$ with two columns interchanged. Or $det(B)=Kdet(A)$ IF $B$ is $A$ with a column multiplied by $K$, there are some few known importants properties of determinantes of linear algebra. It is no requested to use some weird properties. – JuanMuñoz Oct 01 '14 at 22:50
  • @Amzoti I have already use Sarrus Law, but it get to much complicated, in this excersise is just requested to avoid using that law, and get the result in a simpler way using de correct properties of determinants. – JuanMuñoz Oct 01 '14 at 22:59
  • @MattAllegro I just wrote you something after your comment but I forgot to call you, maybe the next comment I did, could help you to indicate a better way to proceed. – JuanMuñoz Oct 01 '14 at 23:03
  • @Amzoti Cofactors and Sarrus, are not what I refer to be Determinant properties. Those are what I need to use. I gues Im in the same hole. – JuanMuñoz Oct 01 '14 at 23:05
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    See also How to solve this determinant. Found using Approach0 – Martin Sleziak Jan 10 '17 at 05:52
  • @MartinSleziak In your link the matrix do not have de factor a^2, in my matrix the factor a^2 is in the first , so I it is not the same excersise. – JuanMuñoz Jan 25 '18 at 20:53

1 Answers1

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the value of the determinant is evidently a homogeneous symmetric polynomial D(a,b,c) of degree six.

it is easy to see that it vanishes if $a=0$, $b=0$ or $c=0$.

and if $a+b+c=0$ then $(b+c)^2=a^2$ etc. , so again the determinant vanishes.

hence for suitably chosen $\lambda$ and $\mu$ $$ D(a,b,c) = abc(a+b+c)\left(\lambda (a^2+b^2+c^2) + \mu(ab+bc+ca) \right) $$ we may easily compute $D(1,1,1)=54$, hence $$ \lambda+\mu = 6 $$ likewise $D(2,1,1) = 256$, giving: $$ 6\lambda +5 \mu = 32 $$thus $\lambda=2$ and $\mu=4$, so $$ \lambda (a^2+b^2+c^2) + \mu(ab+bc+ca) = 2(a+b+c)^2 $$ and, finally: $$ D(a,b,c) = 2abc(a+b+c)^3 $$

David Holden
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    This needs solved like this \begin{equation} \det\begin{pmatrix}1 & 1 & 1 \ x & y & z \ x^2 & y^2 & z^2 \ \end{pmatrix} = \det\begin{pmatrix}1 & 0 & 0 \ x & y-x & z-x \ x^2 & y^2-x^2 & z^2-x^2 \ \end{pmatrix} = \det\begin{pmatrix}y-x & z-x \ y^2-x^2 & z^2-x^2 \ \end{pmatrix} = \det\begin{pmatrix}y-x & z-x \ (y-x)(y+x) & (z-x)(z+x) \ \end{pmatrix}=(y-x)(z-x)\det\begin{pmatrix} 1 & 1 \ y+x & z+x \ \end{pmatrix} = (y-x)(z-x)\det\begin{pmatrix} 1 & 0 \ y+x & z-y \ \end{pmatrix}=(y-x)(z-x)(z-y) \end{equation} – JuanMuñoz Oct 03 '14 at 21:58
  • This past comment reflects how the excercise requires to be @DavidHolden solved, so your answer is very comprehensible, but this excercise needs to be solved by known procedures. Like the past I have posted, but I have really enjoid your answer. – JuanMuñoz Oct 07 '14 at 04:14
  • How can you write $D(a,b,c) = abc(a+b+c)\left[\lambda (a^2+b^2+c^2) + \mu(ab+bc+ca) \right]$ after you obtained $abc(a+b+c)$ as factor ? – SOORAJ SOMAN Feb 28 '18 at 07:56
  • that is the form is must take if it is a homogeneous symmetric polynomial of degree six divisible by $abc(a+b+c)$. – David Holden Feb 28 '18 at 12:03