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The matrix is defined like bellow, for $n=3$,

$$ A = \begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\\ 7 & 8 & 9\end{bmatrix} $$ and it has $\det(A)=0$.

For $n=4$, the matrix $$ \left(\begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \end{array}\right) $$ has determinant equals to $0$ too. So, I check a few and for $n\in\{5,6,7\}$ the determinants are $0$ too.

Do all such matrices have determinant $0$ for $n\ge3$?

Bernard
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Morilla Thaisa
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1 Answers1

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Yes.

Indeed, you can see that $2R_2-R_1-R_3=0$, where $R_i$ denotes the $i$th row. So there is a linear non trivial relation between the rows, so the determinant is $0$.

TheSilverDoe
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    For some perspective for the OP: since the rows are in arithmetic progression, the second discrete derivative (second difference) of three consecutive rows is always zero. Likewise for the columns. This can be extended to higher order differences if the rows are polynomial functions of the row index. – Jean-Claude Arbaut Oct 02 '20 at 15:52
  • Ah, yes. The three consecutive rows. – Morilla Thaisa Oct 02 '20 at 15:59