The value of the determinant

Question

Calculate the value of determinant: $$D = \begin{vmatrix} 0 & 1 & 2 & ... & 2020 \\ 1 & 0 & 1 & ... & 2019 \\ 2 & 1 & 0 & ... & 2018 \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ 2019 & 2018 & 2017 & ... & 1 \\ 2020 & 2019 & 2018 & ... & 0 \\ \end{vmatrix}$$

I tried to change $L_k$ with $L_{n-k}$ and i got a circular determinant but i don't know to solve it.

Answer 1

Sorry, I can't help you. I'm just giving an answer to tell you that I put the matrix on matlab and tried it for different dimensions $N$. Considering: $$D = \begin{vmatrix} 0 & 1 & 2 & ... & N\\ 1 & 0 & 1 & ... & N-1\\ 2 & 1 & 0 & ... & N-2\\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ N-1& N-2& N-3& ... & 1 \\ N& N-1& N-2& ... & 0 \\ \end{vmatrix}$$ The results are huge numbers.

• $N = 20 \qquad \det D = 1.0486e+07$
• $N = 200 \qquad \det D = 1.6069e+62$
• $N = 2020 \qquad \det D = Inf$

In any case the trace resulted in 0 obviously. Hope it will help you, somehow...

It follows the matlab script I used, so you can try and see yourself, that one is very easy.

---

N = 2020;

A = zeros(N,N);

for i = 0:N
    for j = 0:i
        A(i+1,j+1) = abs(j-i);
        A(j+1,i+1) = abs(j-i);
    end
end

N
Trace = trace(A)
Determinant = det(A)

Answer 2

Let's do this for a $5 \times 5$ matrix and hopefully it would make sense how to generalize this. The key idea is that when you apply operations of Gaussian Elimination to the matrix,

• flipping rows multiplies the result by
• rescaling a row by $k$ rescales the determinant by $k$ as well
• adding multiples of rows to other rows does not change the determinant.

So, $$ \begin{split} D &= \begin{vmatrix} 0 & 1 & 2 & 3 \\ 1 & 0 & 1 & 2 \\ 2 & 1 & 0 & 1 \\ 3 & 2 & 1 & 0 \\ \end{vmatrix} = \begin{vmatrix} -1 & 1 & 2 & 3 \\ 1 & 0 & 1 & 2 \\ 1 & 1 & 0 & 1 \\ 1 & 2 & 1 & 0 \\ \end{vmatrix} = \begin{vmatrix} -1 & -1 & 2 & 3 \\ 1 & -1 & 1 & 2 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{vmatrix} = \begin{vmatrix} -1 & -1 & -1 & 3 \\ 1 & -1 & -1 & 2 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & 0 \end{vmatrix} \\ & = \begin{vmatrix} -1 & -1 & -1 & 3 \\ 0 & -2 & -2 & 5 \\ 0 & 0 & -2 & 4 \\ 0 & 0 & 0 & 3 \end{vmatrix} = (-1) \cdot (-2) \cdot (-2) \cdot 3 = 12. \end{split} $$ We subtract column 2 from column 1, and then column 3 from column 2, and then column 4 from column 3. The second row starts by subtracting the first row from every other row to get a diagonal matrix....