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Find $$ \Delta=\begin{vmatrix} \frac{1}{a+x} &\frac{1}{b+x} &\frac{1}{c+x} \\ \frac{1}{a+y} &\frac{1}{b+y} &\frac{1}{c+y} \\ \frac{1}{a+z} &\frac{1}{b+z} &\frac{1}{c+z} \end{vmatrix}$$

I applied $C_1 \to C_1-C_2$ and $C_2 \to C_2-C_3$ we get

$$ \Delta=\begin{vmatrix} \frac{b-a}{(a+x)(b+x)} &\frac{c-b}{(b+x)(c+x)} &\frac{1}{c+x} \\ \frac{b-a}{(a+y)(b+y)} &\frac{c-b}{(b+y)(c+y)}&\frac{1}{c+y} \\ \frac{b-a}{(a+z)(b+z)} &\frac{c-b}{(b+z)(c+z)} &\frac{1}{c+z} \end{vmatrix}$$

Now taking $b-a$,$\:$$c-b$ common and taking $$\frac{1}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)}$$ outside Determinant we get

$$\Delta=\frac{(b-a)(c-b)}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)} \times \begin{vmatrix} c+x &a+x &(a+x)(b+x) \\ c+y &a+y &(a+y)(b+y)\\ c+z &a+z &(a+z)(b+z) \end{vmatrix}$$

Now apply $C_1 \to C_1-C_2$ we get

$$\Delta=\frac{(b-a)(c-b)(c-a)}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)} \times \begin{vmatrix} 1 &a+x &(a+x)(b+x) \\ 1 &a+y &(a+y)(b+y)\\ 1 &a+z &(a+z)(b+z) \end{vmatrix}$$

any clue here?

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  • $\begingroup$ There's a formula for 3x3 determinants, so why not simply use that ? $\endgroup$
    – StephenG
    Jun 12 '17 at 9:42
  • $\begingroup$ Also, google for Cauchy determinant. $\endgroup$ Jun 12 '17 at 10:17
  • $\begingroup$ I am not sure but this matrix looks like a Cauchy-Matrix where we have an explicit formula for the determinant. Maybe this may help. en.m.wikipedia.org/wiki/Cauchy_matrix. Edit: Seems like I was too slow. $\endgroup$
    – Deavor
    Jun 12 '17 at 10:20
  • $\begingroup$ You could add to your post that you are trying to evaluate it using property of determinants, lest anyone mentions that there is the Sarrus's rule or any other method to work with. $\endgroup$ Jun 12 '17 at 11:30
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\begin{align} \Delta&=\begin{vmatrix} \frac{1}{a+x} & \frac{1}{b+x} & \frac{1}{c+x} \\ \frac{1}{a+y} & \frac{1}{b+y} & \frac{1}{c+y} \\ \frac{1}{a+z} & \frac{1}{b+z} & \frac{1}{c+z}\end{vmatrix}\\ \left[\prod_{q\in\{x,y,z\}}(a+q)\right]\Delta&=\begin{vmatrix} 1 & \frac{a+x}{b+x} & \frac{a+x}{c+x} \\ 1 & \frac{a+y}{b+y} & \frac{a+y}{c+y} \\ 1 & \frac{a+z}{b+z} & \frac{a+z}{c+z}\end{vmatrix}\\ \left[\prod_{p\in\{a,b\}}\prod_{q\in\{x,y,z\}}(p+q)\right]\Delta&=\begin{vmatrix} b+x & a+x & \frac{(a+x)(b+x)}{c+x} \\ b+y & a+y & \frac{(a+y)(b+y)}{c+y} \\ b+z & a+z & \frac{(a+z)(b+z)}{c+z}\end{vmatrix}\\ \left[\prod_{p\in\{a,b,c\}}\prod_{q\in\{x,y,z\}}(p+q)\right]\Delta&=\begin{vmatrix} (b+x)(c+x) & (c+x)(a+x) & (a+x)(b+x) \\ (b+y)(c+y) & (c+y)(a+y) & (a+y)(b+y) \\ (b+z)(c+z) & (c+z)(a+z) & (a+z)(b+z)\end{vmatrix} \end{align}

Let $\displaystyle \Delta_0=\begin{vmatrix} (b+x)(c+x) & (c+x)(a+x) & (a+x)(b+x) \\ (b+y)(c+y) & (c+y)(a+y) & (a+y)(b+y) \\ (b+z)(c+z) & (c+z)(a+z) & (a+z)(b+z)\end{vmatrix}$.

Note that $\Delta_0=0$ when $a=b$, $b=c$, $c=a$, $x=y$, $y=z$ and $z=x$.

Since $\Delta_0$ is a polynomial of degree $6$ in $a,b,c,x,y,z$,

$$\Delta_0=k(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$$

for some constant $k$.

Put $a=x=1$, $b=y=0$ and $c=z=-1$.

\begin{align} \begin{vmatrix} 0 & 0 & 2 \\ 0 & -1 & 0 \\ 2 & 0 & 0\end{vmatrix}&=k(1)(1)(-2)(1)(1)(-2)\\ k&=1 \end{align}

Therefore,

$$\Delta=\frac{(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)}{(a+x)(a+y)(a+z)(b+x)(b+y)(b+z)(c+x)(c+y)(c+z)}$$

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