How to prove the determinant of
$$A=(a_{ij})_{n\times n}=b_i^2-b_j^2$$ where $b_1,b_2,...,b_n$ are some distinct real numbers , $n\geq 3$ and $i,j=1,2,...,n$ is zero?
Here, $A$ is skew symmetric and I know odd order skew symmetric matrix has determinant zero. How about the even case? or Is there any other method to show $\det(A)=0$ ?
You can't do it, at least not for all $n$, as the question was originally asked.
Pick $b_1 = 1$ and $b_2 = 0$ to get the matrix $$ \pmatrix{0 & 1 \\ -1 & 0} $$ whose determinant is not zero.