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The following problem is taken from here exercise $2:$

Question: Evaluate the determinant: \begin{vmatrix} 0 & x & x & \dots & x \\ y & 0 & x & \dots & x \\ y & y & 0 & \dots & x \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y & y & y & \dots & 0 \end{vmatrix}

My attempt:

I tried to use first row substract second row to obtain \begin{pmatrix} y & -x & 0 \dots & 0 \end{pmatrix} and also first row subtracts remaining rows. However, I have no idea how to proceed.

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  • $\begingroup$ I assume that you know the dimension of the matrix. Then, operating like you did in order to get the first (or second) row to be all elements zero but two of them, you can use the expansion formula of the definition of determinant to transform it into two determinants of one row less (and apply this again to them until you finish). en.wikipedia.org/wiki/Determinant#Definition $\endgroup$ – Edu Aug 21 '17 at 13:36
  • $\begingroup$ Do you know the Sarrus Rule? It might help. $\endgroup$ – MCCCS Aug 21 '17 at 13:49
  • $\begingroup$ @MCCCS: Yes, I know the Sarrus Rule. But it is only applicable on $3\times 3$ matrix, right? $\endgroup$ – Idonknow Aug 21 '17 at 13:50
  • $\begingroup$ @Idonknow Yes you're right. $\endgroup$ – MCCCS Aug 21 '17 at 13:51
  • $\begingroup$ @MCCCS: However, the matrix in the question may not be $3\times 3.$ How does Sarrus Rule help? $\endgroup$ – Idonknow Aug 21 '17 at 13:52
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If you continue subtracting the $(k+1)$th row from the $k$th, you end up with $$ \Delta_n = \begin{vmatrix} -y & x \\ & -y & x \\ && -y & x \\ &&&\ddots & \ddots \\ &&&& -y & x \\ y & y & y & \cdots & y & 0 \end{vmatrix}, $$ where blank spaces are zeros. One can also pull out a factor of $xy$ now, but there's not a lot of point. Expanding along the first column gives $$ \Delta_n = -y\Delta_{n-1} +(-1)^{n-1} y \begin{vmatrix} x \\ -y & x \\ & -y & x \\ &&\ddots & \ddots \\ &&& -y & x \end{vmatrix} = -y \Delta_{n-1} + (-1)^{n-1} yx^{n-1}. $$ Iterating this and using $\Delta_2=-xy$ gives $$ \Delta_n = (-1)^{n-1} (xy^{n-1}+x^2y^{n-2}+\dotsb+x^{n-1}y) = (-1)^{n-1}xy \frac{x^{n-1}-y^{n-1}}{x-y}. $$

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  • $\begingroup$ I think the last formula should be $(-1)^{n-1} \left(\frac{x^{n+1}-y^{n+1}}{x-y}-x^n-y^n\right)$ or $(-1)^{n-1}\frac{x^n y-y^n x}{x-y}$ $\endgroup$ – Reinhard Meier Aug 21 '17 at 14:37
  • $\begingroup$ Yes, good point. $\endgroup$ – Chappers Aug 21 '17 at 14:52
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Let $A$ be a $n\times n$ matrix of the form described above. You can easily compute the determinant by hand for the case up to $n=4$, which suggests the following relation: $$\det(A_n) = (-1)^{n+1}\sum_{i=1}^{n-1}x^iy^{n-i}$$ This can be proved by induction, by using the calculated small cases, followed by using the fact that the determinant can be computed as the sum of the determinants of comatrices.

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Chappers's expanding is correct, but the answer is imho wrong: it is not correct for $n=2$. Iterating over $\Delta$ gives you $$ \Delta_n = (-1)^{n-1}xy\frac{x^{n-1}-y^{n-1}}{x-y} $$

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