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This works if the integers are listed in consecutive order either along the rows or columns. Why does the determinant of the square matrix always equal $0$ for $ n > 2 $?

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    $\begingroup$ I think you need dimension $n \times n$ where $n \ge 3.$ $\endgroup$ – coffeemath Jan 20 '18 at 4:15
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    $\begingroup$ $\det\left(\begin{bmatrix}1&2\\3&4\end{bmatrix}\right)=-2$ $\endgroup$ – JMoravitz Jan 20 '18 at 4:15
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Assuming that $n\geq 3$ and the integers are placed along the rows:

$$\begin{bmatrix} 1&2&3&\cdots&n\\ n+1&n+2&n+3&\cdots&n+n\\ 2n+1&2n+2&2n+3&\cdots&2n+n\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ (n-1)n+1&(n-1)n+2&(n-1)n+3&\cdots&n^2 \end{bmatrix}$$

Consider the second row minus the first row. It will be $[n~n~n~n~\dots~n]$

Now, consider the third row minus the first row. It will be $[2n~2n~2n~2n~\dots~2n]$. It is a constant multiple of another row in the matrix, and as such the rows are not linearly independent.

Worded another way, letting $R_1,R_2,R_3$ denote the first, second and third rows respectively, we have $R_1-2R_2+R_3=0$, proving that they are linearly dependent on one another.

A matrix has nonzero determinant if and only if its rows are linearly independent. Since the rows are dependent on one another, the determinant must be zero.

Note: This does not work for $n=1$ or $n=2$.

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  • $\begingroup$ The relation $R_1-2R_2+R_3=0$ holds even if the integer sequence does not start from $1$. Phrased differently, all the rows of this matrix are linear combinations of the vectors $(1,2,3,\ldots,n)$ and $(1,1,1,\ldots,1)$, so the rank of the matrix is two. $\endgroup$ – Jyrki Lahtonen Jan 6 '19 at 6:45
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At least for 3x3, I think this is why:

We start with the generic form... $$ \left[\begin{array}{ccc} x & x+1 & x+2 \\ x+3 & x+4 & x+5 \\ x+6 & x+7 & x+8 \end{array}\right] $$ Then subtract the last row from the first... $$ \left[\begin{array}{ccc} x & x+1 & x+2 \\ x+3 & x+4 & x+5 \\ 6 & 6 & 6 \end{array}\right] $$ Then the second from the first... $$ \left[\begin{array}{ccc} x & x+1 & x+2 \\ 3 & 3 & 3 \\ 6 & 6 & 6 \end{array}\right] $$ And we see clearly that the matrix is not full rank.

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For $n\gt 2$, it is easy to see that $R_k=R_{k-1}+n$... Thus by doing a couple row operations (namely subtracting $R_k$ from $R_{k+1}$ for a couple different choices of $k$) we get two rows whose entries are all $n$... then subtract one of these from the other to get a row of zeros...

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First of all it is not always true. For example determinant of $$ \begin{bmatrix} 1&2\\3&4\\ \end{bmatrix} $$

is $-2$

When you have a larger matrix and the rows become linearly dependent due to the arrangement of consecutive numbers then the determinant is zero.

You can see that from elementary row operations.

Note that in general, the sum of the first row and the third row is twice the second row therefore, these three rows are linearly dependent.

For example in $$ \begin{bmatrix} 1&2&3\\4&5&6\\7&8&9 \end{bmatrix} $$ we get $R_1 + R_3 =2R_2$, thus $R_1 + R_3 -2R_2 = [0,0,0]$

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  • $\begingroup$ It would probably be useful to describe what row operations are needed to see that the determinant is zero. Otherwise, your explanation «the determinant is zero because you can generate a row of sería using row operations» applies to every matrix with zero determinant and is therefore not very useful! $\endgroup$ – Mariano Suárez-Álvarez Jan 20 '18 at 4:45
  • $\begingroup$ @MarianoSuárez-Álvarez I have mentioned that multiple reduction of rows results in two equal rows or a row of all zero's. In my example of the $ 3\times 3$ matrix, if you subtract the first row from the second row you get a row of $3$ 's and if you subtract the first row from the third row you get a row of $6$'s, hence the linear dependence. $\endgroup$ – Mohammad Riazi-Kermani Jan 20 '18 at 5:12
  • $\begingroup$ My point is, for every single matrix with zero determinant you can say that you get zero because "multiple reduction of does results in two equal rows or a row of all zeros" so that your answer would be immensely improved if you actually explained in it (and not in a comment) how exactly that happens for these specific matrices. $\endgroup$ – Mariano Suárez-Álvarez Jan 20 '18 at 5:54
  • $\begingroup$ @MarianoSuárez-Álvarez Done! Thanks for your constructive comments. $\endgroup$ – Mohammad Riazi-Kermani Jan 20 '18 at 6:09
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Alternatively, using arithmetic mean, for $3\times 3$ matrix determinant: $$\begin{vmatrix} x+1 & x+2 & x+3 \\ x+4 & x+5 & x+6 \\ x+7 & x+8 & x+9 \end{vmatrix}=0,$$ because: $C_1+C_3=2C_2$, which shows linear dependence of the column vectors.

For $4\times 4$ matrix determinant: $$\begin{vmatrix} x+1 & x+2 & x+3 & x+4 \\ x+5 & x+6 & x+7 & x+8 \\ x+9 & x+10 & x+11 & x+12 \end{vmatrix}=0, \ \text{because: } C_1+C_4=C_2+C_3.$$

In geneal, for $(2n+1)\times (2n+1), n>0,$ matrix determinant: $$\begin{vmatrix} x+1&\cdots &x+(n+1) &\cdots &x+(2n+1)\\ x+(2n+1)+1&\cdots&x+(2n+1)+n+1&\cdots&x+2(2n+1)\\ \vdots&\vdots&\vdots &\vdots&\vdots\\ x+2n(2n+1)+1&\cdots &x+2n(2n+1)+n+1&\cdots&x+(2n+1)^2 \end{vmatrix}=0,\\ \text{because: } C_1+C_{2n+1}=C_{n+1}.$$ Can you write the generalization for $(2n)\times (2n),n>0,$ matrix determinant?

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