Let $n$ be a natural number and $a,b,c\in\Bbb R$.

How to calculate following determinant?

$$ \begin{vmatrix} a & b & b & b & \dots & b & b \\ c & a & b & b & \dots & b & b \\ c & c & a & b & \dots & b & b \\ \vdots & \ddots&\ddots& \ddots &\ddots &\ddots& \vdots \\ c & c & c & c &\dots & a & b \\ c & c & c & c & \dots & c & a \end{vmatrix} $$

I. e. matrix has only $a$s in the diagonal, only $b$s in the top right and only $c$s in the bottom left.

I tried to develop the first row/column but te calculations don't lead anywhere. Please give me some help.

  • $\begingroup$ Welcome to Maths SX! Is $b\ne c$? $\endgroup$
    – Bernard
    Sep 26 '19 at 20:38
  • $\begingroup$ This is a Toeplitz matrix. If it is symmetric or Hermitian, it can be easier to find some formulas. $\endgroup$
    – user376343
    Sep 26 '19 at 20:42
  • $\begingroup$ @Bernard Yes, let's suppose $b\neq c$ $\endgroup$
    – user708986
    Sep 26 '19 at 21:10

Subtract the 2nd column from the 1st, the 3rd from the 2nd, the 4th from the 3rd, &c. We obtain the determinant $$D_n=\begin{vmatrix} a-b&0&0&0&\dots &0&b\\ c-a& a-b&0&0&\dots &0&b \\ 0&c-a& a-b&0& \dots &0&b \\[-1ex] \vdots &&&\ddots&&\vdots\\ 0&0&0&0&\dots&a-b&b \\ 0&0&0&0&\dots&c-a&a \end{vmatrix}$$ Expand along the 1st row, noting the $(1,1)$ cofactor is just $D_{n-1}$ and the $(1,n)$ cofactor is upper triangular: $$D_n =(a-b)D_{n-1}+(-1)^{n-1}b(c-a)^{n-1}=(a-b)D_{n-1}+b(a-c)^{n-1}$$ On the other hand, swapping $b$ and $c$ changes the original matrix into its transpose. So in the above relation, we can swap $b$ and $c$, and $D_n$ also satisfies the relation $$D_n=(a-c)D_{n-1}+c(a-b)^{n-1},$$ whence by subtraction, $(b-c)D_{n-1}=b(a-c)^{n-1}-c(a-b)^{n-1}$, and ultimately $$D_{n-1}=\frac{b(a-c)^{n-1}-c(a-b)^{n-1}}{b-c}\qquad\text{ if }b\ne c$$

Case $\;b=c\,$:

Consider the numerator of $D_n$, for the case $b\ne c$ as a function of $x=c$: $$f(x)=b(a-x)^n-x(a-b)^n.$$ Note that $f(b)=0$, so that $D_n$ is $$D_n=-\frac{f(c)-f(b)}{c-b},\quad\text{ which tends to }\; -f'(b)\;\text{ when }\;c\to b.$$ So by continuity of the determinant,in the case $b=c$, we have \begin{align}D_n &=nb(a-x)^{n-1}+(a-b)^n\bigg|_{x=b}=nb(a-b)^{n-1}+(a-b)^n\\[1ex] &=(a-b)^{n-1}\bigl(a+(n-1)b\bigr). \end{align}


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