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$$ D = \begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 2 & 1 & 1 & \dots & 1 & 0 \\ 3 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ n-1 & 1 & 0 & \dots & 0 & 0 \\ n & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix} =n*1*(-1)^\frac{n(n-1)}{2} $$
I don't quite understand the solution of this determinant. I do understand that if we use Laplace expansion along the last row we get $$ D = n* \begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix} $$ But how does the remaining determinant euqal: $1*(-1)^\frac{n(n-1)}{2}$?



Edit:

$$ \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{vmatrix} =(-1)^{4} \begin{vmatrix} 0 & 1 \\ 1 & 0 \\ \end{vmatrix} =(-1)^{4+3+2} $$ I thought it should go: $(-1)^{3+2+1}$ or is the power actually the sum of row and column coordinates?

A6EE
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    When you expand about the bottom row you need to take into account the sign which is $(-1)^n$. The full determinant therefore becomes $n(-1)^{n + (n-1) + \ldots +2 + 1}$ – Winther Dec 10 '15 at 10:54
  • Oh my God, thank you! – A6EE Dec 10 '15 at 10:56
  • @Winther One more question, I don't understand the sign. My teacher said the sign in this case is $(-1)^{n+1}$. She did an example for a $3x3$ determinant. I'll post it in the edit section. I'm not sure to what power should $-1$ be. – A6EE Dec 10 '15 at 11:01
  • Yes the power is the sum of the row and column indices. – KittyL Dec 10 '15 at 11:10
  • You are right: it's $(-1)^{n+1}$, not $(-1)^{n}$ as I said above! – Winther Dec 10 '15 at 11:11

2 Answers2

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To compute

$$\begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix}$$

You can proceed like that: If we swap the $k$-th row with the $n-k$-th row of the matrix for $k=1,2,\ldots$ then, at some point, we will get a lower triangular matrix with only $1$ on its main diagonal (so its determinant is $1$). Now, each swap corresponds to an elementary operation of the Gauss-Jordan method and thus changes the sign of the determinant (i.e. multiplies it by $-1$). So count the number of swaps to get the result.

Martin Sleziak
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Surb
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First do an elementary transformation that does not change the determinant, namely add to the first column the sum of all the others. You are left with $$ \begin{vmatrix} n & 1 & 1 & \dots & 1 & 1 \\ n & 1 & 1 & \dots & 1 & 0 \\ n & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ n & 1 & 0 & \dots & 0 & 0 \\ n & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix} = n \cdot \begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix} = n \cdot (-1)^{s}. $$ Here $s$ is the sign of the permutation $\sigma$ on $\{1, 2, \dots, n\}$ that swaps $i$ with $n-i$, since exchanging the $i$-th row with the $(n-i)$-th row of $$ \begin{bmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{bmatrix}, $$ for $1 \le i \le n/2$, we get a matrix which has visibly determinant $1$.

The permutation $\sigma$ has $s = n/2$ two-cycles if $n$ is even, and $s = (n-1)/2$ two-cycles if $n$ is odd. So indeed $$ (-1)^{s} = (-1)^{n (n-1)/2} = \begin{cases} (-1)^{n/2} & \text{if $n$ is even, as $(-1)^{n-1} = -1$,}\\ (-1)^{(n-1)/2} & \text{if $n$ is odd, as $(-1)^{n} = -1$.}\\ \end{cases} $$

Andreas Caranti
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