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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.

LU decomposition

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    $\begingroup$ Hint: the sequence of squares satisfies $k^2 - 3(k+1)^2 + 3(k+2)^2-(k+3)^2 = 0$. $\endgroup$
    – GEdgar
    Jan 17 at 12:35
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    $\begingroup$ Hint, If you subtract the first row from the second you get the sequence of odd integers, Look at other adjacent rows. $\endgroup$ Jan 17 at 12:39
  • $\begingroup$ It seems like a circulant matrix. $\endgroup$
    – Lelouch
    Jan 17 at 13:11
  • $\begingroup$ @Moo Yes, I saw it. Thanks. $\endgroup$
    – jackson
    Jan 17 at 13:42
  • $\begingroup$ @Lelouch are you sure? $\endgroup$
    – DatBoi
    Jan 17 at 14:08

3 Answers 3

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$$A = \det\begin{pmatrix}1&4&9&16&25&36\\4&9&16&25&36&49\\9&16&25&36&49&64\\16&25&36&49&64&81\\25&36&49&64&81&100\\36&49&64&81&100&121\end{pmatrix}$$ By successively subtracting rows, we get

$R_5=R_5-R_4$, $R_4=R_4-R_3$, $R_3=R_3-R_2$, $R_2=R_2-R_4$,

$$A = \det\begin{pmatrix}1&4&9&16&25&36\\7&9&11&13&15&17\\9&11&13&15&17&19\\11&13&15&17&19&21\\13&15&17&19&21&23\\15&17&19&21&23&25\end{pmatrix}$$ We repeat the same again, we get $R_5=R_5-R_4$, $R_4=R_4-R_3$

$$A = \det\begin{pmatrix}1&4&9&16&25&36\\7&9&11&13&15&17\\9&11&13&15&17&19\\11&13&15&17&19&21\\2&2&2&2&2&2\\2&2&2&2&2&2\end{pmatrix}$$

As $R_5=R_4$, by properties of determinants we conclude that the determinant is zero or $A = 0$

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    $\begingroup$ In the very last line, it says "det(A) = 0", but before you had introduced A as the number $A = \det \dots$. $\endgroup$
    – NerdOnTour
    Jan 17 at 15:51
  • $\begingroup$ My fellow told me just like you wrote. Thanks. $\endgroup$
    – jackson
    Jan 17 at 16:55
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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$~3$).

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Let us give your matrix a name:

$$ A= \begin{pmatrix} 1&4&9&16&25&36\\4&9&16&25&36&49\\9&16&25&36&49&64\\16&25&36&49&64&81\\25&36&49&64&81&100\\36&49&64&81&100&121 \end{pmatrix} $$

The decomposition $A = LU$ (which you have found explicitly) is very helpful in this case. The upper right matrix ($U$) does not have full (row) rank, thus its determinant is $0$. Since the determinant is a multiplicative function, we get

$$ \det A = \det(L) \cdot \det(U) = \det(L) \cdot 0 = 0 $$

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