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Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$

There must be some trick, like using elementary row operations, to get the determinant into that form, but I am not seeing it. And directly computing the determinant by the cofactor expansion looks very nasty. So is there a simpler way to compute this determinant?

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  • $\begingroup$ With the rule of Sarrus it shouldn't be too hard and the matrix on the left hand side looks very symmetric. $\endgroup$
    – Peter
    Dec 28, 2019 at 12:07
  • $\begingroup$ Multiply the right matrix with itself and use $\det(A*B)=\det(A) * \det(B)$ $\endgroup$
    – Peter
    Dec 28, 2019 at 12:11

4 Answers 4

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Use the rule of Sarrus to show $\begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = (2abc)^2$ and then show that $A^2 :=\begin{pmatrix}0&c&b\\c&0&a\\b&a&0\end{pmatrix}^2=\begin{pmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{pmatrix}$ and then use $\det(A*A)=\det(A)*\det(A)$.

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  • $\begingroup$ en.wikipedia.org/wiki/Rule_of_Sarrus $\endgroup$
    – Peter
    Dec 28, 2019 at 12:18
  • $\begingroup$ I see, so $\det A:=\begin{pmatrix}0&c&b\\c&0&a\\b&a&0\end{pmatrix}^2$ = $\det A\det A = $(\det A)^2$, from which the result follows. $\endgroup$
    – Math1000
    Dec 28, 2019 at 14:48
  • $\begingroup$ Your comment cannot be true on the left side you have a number and on the right hand side a 3 times 3 matrix! $\endgroup$
    – Peter
    Dec 28, 2019 at 19:15
  • $\begingroup$ If you were satisfied with the answer please accept it as answer:) $\endgroup$
    – Peter
    Dec 28, 2019 at 19:16
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A determinant of the left matrix is equal to $$\prod_{cyc}(a^2+b^2)+2a^2b^2c^2-\sum_{cyc}(a^2+b^2)a^2b^2=$$ $$=\sum_{cyc}\left(a^4b^2+a^4c^2+\frac{2}{3}a^2b^2c^2\right)+2a^2b^2c^2-\sum_{cyc}(a^4b^2+a^4c^2)=4a^2b^2c^2.$$ A determinant of the right matrix is equal to $$0+2abc-0=2abc$$ and we are done!

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Hint:

$$\triangle=\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \dfrac1a\begin{vmatrix}a(b^2+c^2)&ab&ac\\a(ba)&c^2+a^2&bc\\a(ca)&cb&a^2+b^2\end{vmatrix} $$

$$C_1'=C_1-bC_2-cC_3$$

$$\triangle=\dfrac1a\begin{vmatrix}0&ab&ac\\-2bc^2&c^2+a^2&bc\\-2bc^2&cb&a^2+b^2\end{vmatrix}=-2bc\begin{vmatrix}0&b&c\\c&c^2+a^2&bc\\b&cb&a^2+b^2\end{vmatrix}=2bc\begin{vmatrix}0&c&b\\c&bc&c^2+a^2\\b&a^2+b^2&cb\end{vmatrix}$$

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

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

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Let $b=ax,c=axy$. Then: $$\small\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} =\begin{vmatrix}a^2x^2(1+y^2)&a^2x&a^2xy\\a^2x&a^2(1+x^2y^2)&a^2x^2y\\a^2xy&a^2x^2y&a^2(1+x^2)\end{vmatrix} =a^6x^2\begin{vmatrix}1+y^2&1&y\\1&1+x^2y^2&x^2y\\y&x^2y&1+x^2\end{vmatrix} =\\ \small\stackrel{R_1-R_2\to R_1}= a^6x^2\begin{vmatrix}y^2&-x^2y^2&y-x^2y\\1&1+x^2y^2&x^2y\\y&x^2y&1+x^2\end{vmatrix} =a^6x^2y\begin{vmatrix}y&-x^2y&1-x^2\\1&1+x^2y^2&x^2y\\y&x^2y&1+x^2\end{vmatrix} \stackrel{R_3-R_1\to R_3}=\\ \small a^6x^2y\begin{vmatrix}y&-x^2y&1-x^2\\1&1+x^2y^2&x^2y\\0&2x^2y&2x^2\end{vmatrix}= 2a^6x^4y\begin{vmatrix}y&-x^2y&1-x^2\\1&1+x^2y^2&x^2y\\0&y&1\end{vmatrix}=\\ \small =2a^6x^4y\cdot [y(1+x^2y^2-x^2y^2)-(-x^2y-y(1-x^2)]=4a^6x^4y^2=4a^2b^2c^2.$$

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