Is there a more efficient way to calculate the determinant of this matrix? I used Gaussian Elimination to calculate $$\det\begin{pmatrix}1&4&9&16&25&36\\4&9&16&25&36&49\\9&16&25&36&49&64\\16&25&36&49&64&81\\25&36&49&64&81&100\\36&49&64&81&100&121\end{pmatrix}$$ and found the answer to be $0$. It took a lot of time to do.
The lower-upper (LU) decomposition is shown below. I think that there might be a more efficient way to calculate the determinant of this kind of matrix.


Clearly every row is part of the space of vectors $(a_1,\ldots,a_6)$ for which $a_i$ can be given by a polynomial expression in$~i$ of degree${}<3$ (i.e., with $a_i=p+qi+ri^2$ for some scalars $p,q,r$ and $0<i\leq 6$). That subspace of $\Bbb Q^6$ being of dimension$~3$, any $4$ or more rows are linearly dependent, so the determinant of the matrix must be$~0$ (and the rank of the matrix at most$~3$).
