Calculating the determinant of $A$ with $A_{ij}=a$ for $i<j$, $A_{ij}=-a$ for $i>j$, $A_{ii}=x$, using a pen and paper

Question

Let

$$A = \left[\begin{array}{cccccc} x&a&a&a&\dotsm&a\\ -a&x&a&a&\dotsm&a\\ -a&-a&x&a&\dotsm&a\\ -a&-a&-a&x&\dotsm&a\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ -a&-a&-a&-a&\dotsm&x \end{array}\right].$$

How would one calculate the determinany of this matrix using a pen and paper? Using cofactors is rather computationally intensive; there must be some way to take advantage of the symmetry in this problem.

Answer 1

You may find the determinant by using elementary row operations. Subtract, simultaneously, the first row from the second row, the second row from the third row, etc., we can reduce $A$ to $$ \left[\begin{array}{cccccc} x&a&a&\cdots&\cdots&\cdots&a\\ -x-a&x-a&0&0&\dots&\cdots&0\\ 0&-x-a&x-a&0&\ddots&\ddots&0\\ 0&0&-x-a&x-a&\ddots&\ddots&0\\ \vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&-x-a&x-a&0\\ 0&0&\cdots&\cdots&0&-x-a&x-a \end{array}\right]. $$ So, if we define $D_n$ to be the determinant of $A$ when its size $n\times n$, then by Laplace expansion along the last column, you can obtain a certain recurrence relation for $D_n$. Solving it, you should get $$ D_n = \frac{(x+a)^n + (x-a)^n}2. $$

Answer 2

Hint 1. Try to find eigenvalues.

Hint 2. If your matrix is anti-symmetric then what can you say about eigenvalues?