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If $x+y+z=0$, then prove that $$ \begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix}=xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix} $$

I can do it by Sarrus' law but how can I prove it by matrix operations without actually expanding the determinant ?

My Attempt $$ \begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix}=xyz\begin{vmatrix} a&\frac{yb}{x}&\frac{zc}{x}\\ c&\frac{za}{y}&\frac{xb}{y}\\ b&\frac{xc}{z}&\frac{ya}{z}\\ \end{vmatrix}=xyz\begin{vmatrix} a&-b-\frac{zb}{x}&-c-\frac{yc}{x}\\ c&-a-\frac{xa}{y}&-b-\frac{zb}{y}\\ b&-c-\frac{yc}{z}&-a-\frac{xa}{z}\\ \end{vmatrix}=xyz\begin{vmatrix} a&b+\frac{zb}{x}&c+\frac{yc}{x}\\ c&a+\frac{xa}{y}&b+\frac{zb}{y}\\ b&c+\frac{yc}{z}&a+\frac{xa}{z}\\ \end{vmatrix}=xyz\begin{vmatrix} a&b&c+\frac{yc}{x}\\ c&a&b+\frac{zb}{y}\\ b&c&a+\frac{xa}{z}\\ \end{vmatrix}+xyz\begin{vmatrix} a&\frac{zb}{x}&c+\frac{yc}{x}\\ c&\frac{xa}{y}&b+\frac{zb}{y}\\ b&\frac{yc}{z}&a+\frac{xa}{z}\\ \end{vmatrix}\\ =xyz\bigg(\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}+\begin{vmatrix} a&b&\frac{yc}{x}\\ c&a&\frac{zb}{y}\\ b&c&\frac{xa}{z}\\ \end{vmatrix}+\begin{vmatrix} a&\frac{zb}{x}&c\\ c&\frac{xa}{y}&b\\ b&\frac{yc}{z}&a\\ \end{vmatrix}+\begin{vmatrix} a&\frac{zb}{x}&\frac{yc}{x}\\ c&\frac{xa}{y}&\frac{zb}{y}\\ b&\frac{yc}{z}&\frac{xa}{z}\\ \end{vmatrix}\bigg)\\ $$ I need to prove that the sum of last three terms is zero. $$ \begin{vmatrix} a&b&\frac{yc}{x}\\ c&a&\frac{zb}{y}\\ b&c&\frac{xa}{z}\\ \end{vmatrix}+\begin{vmatrix} a&\frac{zb}{x}&c\\ c&\frac{xa}{y}&b\\ b&\frac{yc}{z}&a\\ \end{vmatrix}+\begin{vmatrix} a&\frac{zb}{x}&\frac{yc}{x}\\ c&\frac{xa}{y}&\frac{zb}{y}\\ b&\frac{yc}{z}&\frac{xa}{z}\\ \end{vmatrix}=\begin{vmatrix} a&b&\frac{yc}{x}\\ c&a&\frac{zb}{y}\\ b&c&\frac{xa}{z}\\ \end{vmatrix}+\begin{vmatrix} a&\frac{zb}{x}&\frac{-zc}{x}\\ c&\frac{xa}{y}&\frac{-xb}{y}\\ b&\frac{yc}{z}&\frac{-ya}{z}\\ \end{vmatrix}\\ $$

Solution by expansion

$$ \Delta=\begin{matrix} xa&yb&zc&xa&yb\\ yc&za&xb&yc&za\\ zb&xc&ya&zb&xc\\ \end{matrix}=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3) $$ We have $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$ as $x+y+z=0$. Thus, $x^3+y^3+z^3=3xyz$ $$ \Delta=xyz(a^3+b^3+c^3)-abc(3xyz)=xyz(a^3+b^3+c^3-3abc)\\ =xyz\bigg[ a\big(a^2-bc\big)-b\big(ac-b^2\big)+c\big(c^2-ab\big) \bigg]\\ =xyz.\bigg[a\begin{vmatrix} a&b\\ c&a\\ \end{vmatrix}-b\begin{vmatrix} c&b\\ b&a\\ \end{vmatrix}+c\begin{vmatrix} c&a\\ b&c\\ \end{vmatrix}\bigg] =xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix} $$

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  • $\begingroup$ Do you need something like $a+b+c=0$ as well? $\endgroup$ – Michael Burr Mar 4 '18 at 13:21
  • $\begingroup$ Perhaps you'll get an answer, but I don't think the solution would be simpler than using Sarrus' rule. If you use Sarrus' rule, the only additional trick is to use the identity $x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(z+x)$, which I don't think is difficult. $\endgroup$ – user1551 Mar 4 '18 at 13:22
  • $\begingroup$ @MichaelBurr nope. just $x+y+z=0$ $\endgroup$ – ss1729 Mar 4 '18 at 13:23
  • $\begingroup$ @KingTut just a shortcut to find determinant en.wikipedia.org/wiki/Rule_of_Sarrus $\endgroup$ – ss1729 Mar 4 '18 at 13:36
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Possible direction, though not a proof.

Do note that curiously, the identity is only true for $n=3$, in higher dimensions it doesn't seem to work. Note also that the matrices do not actually have to be positive semi-definite matrices.

This result may be related to Oppenheim's inequality for Hadamard products, namely if $A,X$: $$A=\begin{pmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{pmatrix},\quad X=\begin{pmatrix} x&y&z\\ y&z&x\\ z&x&y\\ \end{pmatrix}$$ Are two positive semi definite matrices, then: $$\det(A\circ X)\ge (\det A)\prod_i x_{ii}$$

With equality if $A\circ X$ is singular. In your case, you have a circulant and anti-circulant matrices $A,X$ and the above theorem states that: $$\det(A\circ X)\ge xyz\det(A)$$ $$\begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix} \ge xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}$$ You are looking for equality and we of course don't know that either $A$ or $X$ is actually PSD. Moreover we haven't used the condition $x+y+z=0$. Nevertheless, the overall form direction may be one worth pursuing.

Further Reading:
The equality cases for the inequalities of Oppenheim and Schur for positive semi-definite matrices, Czechoslovak Mathematical Journal, Vol. 59 (2009), No. 1, 197–206

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  • $\begingroup$ Presumably you mean $A \circ X$ is singular? $\endgroup$ – preferred_anon Mar 4 '18 at 15:07
  • $\begingroup$ @DanielLittlewood - Thanks, fixed. $\endgroup$ – nbubis Mar 4 '18 at 15:09
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I don't think it is possible without expanding the determinants. The determinant on the right is $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$. The determinant on the left is $(a^3+b^3+c^3)xyz-abc(x^3+y^3+z^3)$. Subtracting the left side from the right side is equal to $abc(x^3+y^3+z^3-3xyz)=abc(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$. The result follows.

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