How to calculate the determinant?

Question

I need help with this determinant ($n \times n)$:

$D_n = \begin{vmatrix} a & x & x & \dots & x & x \\ y & a & x & \dots & x & x \\ y & y & a & \dots & x & x \\ \vdots \\ y & y & y & \dots & a & x \\ y & y & y & \dots & y & a \\ \end{vmatrix}$

I tried to simplify it and derive a reсcurrent formula using Laplace expansion, but it was unsuccessful. Also I had an idea to express the determinant as a sum of determinants.

The answer is: $\displaystyle D_n = \frac{x(a-y)^n - y(a-x)^n}{x-y}$.

Answer 1

Subtract the 2nd column from the first, the 3rd from the second, etc.

Here is what you get in the $4\times4$ case:

$$D_4 = \begin{vmatrix} a-x & 0 & 0 & x \\ y-a & a-x & 0 & x \\ 0 & y-a & a-x & x \\ 0 & 0 & y-a & a \end{vmatrix}$$

Laplace expansion on the first column will yield:

$$D_4=(a-x)D_3-(y-a) \begin{vmatrix} 0 & 0 & x \\ y-a & a-x & x \\ 0 & y-a & a \end{vmatrix}$$

Then again Laplace expansion, on the first row:

$$D_4=(a-x)D_3-(y-a)x \begin{vmatrix} y-a & a-x \\ 0 & y-a \end{vmatrix}=(a-x)D_3-x(y-a)^3$$

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General case: $D_n=(a-x)D_{n-1}-(-1)^nx(y-a)^{n-1}=(a-x)D_{n-1}+x(a-y)^{n-1}$, with $D_1=a$.

Then

$$\begin{align} D_n & = &(a-x)^2D_{n-2}+(a-x)x(a-y)^{n-2}+x(a-y)^{n-1}\\ & = &(a-x)^3D_{n-3}+(a-x)^2x(a-y)^{n-3}+(a-x)x(a-y)^{n-2}+x(a-y)^{n-1}\\ & \vdots & \\ & = &(a-x)^{n-1}D_1+(a-x)^{n-2}x(a-y)+\dots+x(a-y)^{n-1}\\ \end{align}$$

Or

$$D_n=a(a-x)^{n-1}+x\sum_{k=0}^{n-2}(a-x)^{k}(a-y)^{n-1-k}$$

(if you don't like the dots, you can prove the previous formula by induction on $n$)

$$D_n=a(a-x)^{n-1}+x\left(\sum_{k=0}^{n-1}(a-x)^{k}(a-y)^{n-1-k}-(a-x)^{n-1}\right)$$

And, if $u\neq v$,

$$\sum_{k=0}^{n-1}u^{k}v^{n-1-k}=\frac{u^n-v^n}{u-v}$$

Hence, if $x\neq y$,

$$\begin{align} D_n&=a(a-x)^{n-1}-x(a-x)^{n-1}+x\frac{(a-x)^n-(a-y)^n}{(a-x)-(a-y)}\\ &=(a-x)^{n}-x\frac{(a-x)^n-(a-y)^n}{x-y}\\ &=\frac{(x-y)(a-x)^{n}-x(a-x)^n+x(a-y)^n}{x-y}\\ &=\frac{x(a-y)^n-y(a-x)^{n}}{x-y} \end{align} $$

And if $x=y$, you take again the formula

$$\begin{align} D_n&=a(a-x)^{n-1}+x\sum_{k=0}^{n-2}(a-x)^{k}(a-y)^{n-1-k}\\ &=a(a-x)^{n-1}+x(n-1)(a-x)^{n-1}\\ &=(a+(n-1)x)(a-x)^{n-1} \end{align}$$

Answer 2

Here is a short way, if $x\ne y$: first, in $D_{n+1}$, subtract the 2nd column from the 1st, the 3rd from the 2nd, and so on until dead. You get: $$D_{n+1}=\begin{vmatrix} a-x & 0&0&0&\dots&0&x\\ y-a&a-x&0&0&\dots&0&x\\ 0&y-a&a-x&0&\dots&0&x \\ \vdots&&&\ddots&&&\vdots\\ 0&0&0&0&\dots&a-x&x\\ 0&0&0&0&\dots&y-a&a \end{vmatrix}$$ Expand by the first row: you obtain $$D_{n+1}=(a-x)D_n+(-1)^nx(y-a)^n=(a-x)D_n+x(a-y)^n.$$ Now, observe that $\,D_n={}^{\mathrm t\mkern-1mu}D_n$, so $$(a-x)D_n+x(a-y)^n=(a-y)D_n+y(a-x)^n,$$ whence $$(x-y)D_n=x(a-y)^n-y(a-x)^n.$$