# Calculate determinant of matrix with $a$ in top right, $b$ in diagonal and $c$ in bottom left

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.

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