Calculating determinant of a symmetric matrix where the $k$th row is given by $[a_{k-1},a_k,...,a_0,a_1,...,a_{n-(k-1)}]$

Question

For $j = 0,...,n$ set $a_{j} = a_{0} + jd$, where $a_{0}, d$ are fixed real numbers. Calculate the determinant of the $(n+1)\times (n+1)$ matrix $$A = \begin{pmatrix} a_{0} & a_{1} & a_{2} & \dots & a_{n}\\ a_{1} & a_{0} & a_{1} & \dots & a_{n-1}\\ a_{2} & a_{1} & a_{0} & \dots & a_{n-2}\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ a_{n} & a_{n-1} & a_{n-2} & \dots & a_{0} \end{pmatrix}.$$

How to calculate that? I haven't found any property of determinant of symmetric matrix which could help. I've tried to use Gaussian elimination (subtracting each row from the row above it), but it didn't work

Gaussian elimination(subtracting each row from the row above it) brings to the matrix:

$$\begin{pmatrix} -d & d & d & ... & d\\ -d & -d & d & ... & d\\ -d & -d & -d & .... & d\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ a_{n} & a_{n-1} & a_{n-2} & ... & a_{0} \end{pmatrix} = d^{n-1} \cdot \begin{pmatrix} -1 & 1 & 1 & ... & 1\\ -1 & -1 & 1 & ... & 1\\ -1 & -1 & -1 & .... & 1\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ a_{n} & a_{n-1} & a_{n-2} & ... & a_{0} \end{pmatrix}$$

Answer 1

Subtracting from each row the one above it, we shall obtain

$$ \begin{pmatrix} a_{0} & a_{1} & a_{2} & ... & a_{n}\\ d & -d & -d & ... & -d\\ d & d & -d & .... & -d\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ d & d & d & ... & -d \end{pmatrix}$$

Now, subtracting from each column the one before it, we shall obtain

$$ \begin{pmatrix} a_{0} & d & d & ... & d\\ d & -2d & 0 & ... & 0\\ d & 0 & -2d & .... & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ d & 0 & 0 & ... & -2d \end{pmatrix}$$

Next, multiplying columns 2,3,$\ldots$,n+1 by $\frac{1}{2}$ and adding to the first we obtain:

$$ \begin{pmatrix} a_{0}+\frac{nd}{2} & d & d & ... & d\\ 0 & -2d & 0 & ... & 0\\ 0 & 0 & -2d & .... & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ 0 & 0 & 0 & ... & -2d \end{pmatrix}$$

Finally, these row and column operations don't change the determinat of your matrix. Therefore $\det(A)=(a_o+\frac{nd}{2})(-2d)^n$.

Answer 2

Partial Answer: If we use the matrix determinant lemma, then it suffices to consider the case of $a_0 = 0$ and $d = 1$.

With that said, let $A_n$ denote the matrix defined with your formula, with $a_0 = 0$ and $d = 1$. Let $R$ denote an $n \times n$ identity matrix, where the first column has been replaced by $(1,0,\dots,0,n,1-n)$. Note that $$ RA_nR^T = \pmatrix{-2n(n-1) & 0 & 2n\\ 0 & A_{n-1} &-\\ 2n & |}. $$ Conclude that $\det(A_n) = -2n(n-1)\det(A_{n-1}) + 4n^2 \det(A_{n-2})$.