# 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.

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

$$A = \det\begin{pmatrix}1&4&9&16&25&36\\4&9&16&25&36&49\\9&16&25&36&49&64\\16&25&36&49&64&81\\25&36&49&64&81&100\\36&49&64&81&100&121\end{pmatrix}$$ 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 $$A = 0$$

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

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 $$0). That subspace of $$\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$$).

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

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