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Find the determinant of order $n+1$:

$D_n=\begin{vmatrix} 0&1&0&0&\ldots&0&0\\ 1&0&2&0&\ldots&0&0\\ 0&2&0&3&\ldots&0&0\\ 0&0&3&0&\ldots&0&0\\ \vdots &\vdots &\vdots &\vdots &\ddots &\vdots &\vdots\\ 0&0&0&0&\ldots&0&n\\ 0&0&0&0&\ldots&n&0 \end{vmatrix}$

I've tried developing it using Laplace's method, adding rows/columns, but it wasn't very helpful. Please give me some hints, thank you!

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This is a case of a tridiagonal determinant. Expanding it along the last row (or column) you can prove the recurrence relation: $$D_n=-n^2D_{n-2}.$$ From this relation, you can deduce that $$D_n=\begin{cases}(-1)^{\tfrac{n+1}2}(1\cdot 3\cdot \dotsm \cdot n)^2&\text{if $n$ is odd,}\\0&\text{if $n$ is even.}\end{cases}$$

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It seems in this particular case it is strangely feasible to just apply the sum-expression of the determinant (referred to as the Leibniz formula here: https://en.wikipedia.org/wiki/Determinant#n_.C3.97_n_matrices) and directly compute this one.

A nonzero term in the determinant must contain both the ones (1) in the matrix (can you see why?). This eliminates the twos from appearing in a term since they lie in the same column or row as a one, and this forces the threes to be in the term. You can proceed similarly in two interleaved zig-zag patterns down the diagonal to see that all even numbers are eliminated from ever appearing in a nonzero term in the determinant and that all the odd entries must appear.

If $n$ is even, this means that the last column and row can't be covered by any nonzero term of the determinant, so the determinant will be zero, and if $n$ is odd, the determinant will end up being nonzero since all rows and columns are covered, and the determinant then consists of a single term equal in magnitude to the product of the squares of all odd numbers in $[1,n]$. The sign is determined by the parity of the permutation $(1,2)(3,4)\ldots(n,n+1)$, but this permutation is thankfully already factorized into $(n+1)/2$ transpositions.

So summarizing, the determinant is:

  • $0$, if $n$ is even,
  • $(-1)^{(n+1)/2}\prod_{i=0}^{(n-1)/2} (2i+1)^2$, if $n$ is odd.

The upper limit of the above product is carefully set so that we end with $i=n$, which is why the minus sign is there.

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My answer is similar to the one given by A.Sh, but with some differences.

Let $A$ be the given matrix.

Note that $A$ has only one generalized diagonal with nonzero product, namely the diagonal $D$ consisting of all cells whose value is an odd positive integer.

It follows that $\text{det}(A) = \text{det}(B)$ where $B$ is the matrix obtained from $A$ by setting all cells not on $D$ to zero.

We proceed to find $\text{det}(B)$ . . .

First suppose $n$ is even.

Then the last row of $B$ is zero, hence $\text{det}(B) = 0$.

Next suppose $n$ is odd.

Then $B$ is a block diagonal matrix with $2\,{\times}\,2$ square blocks.

It follows that $\text{det}(B)$ is the product of the determinants of those $2\,{\times}\,2$ matrices, hence

\begin{align*} \text{det}(B) &=(-1^2)(-(3^2)\cdots (-(n^2))\\[6pt] &=(-1)^{(n+1)/2}(1\cdot 3\cdots n)^2\\[6pt] \end{align*}

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