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Compute the determinant of the $2n-\text{th}$ order.

$$\begin{vmatrix}0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\3&0&\ldots&0&0&0&0&\ldots&0&2\\2&0&\ldots&0&0&0&0&\ldots&0&3\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\end{vmatrix}$$

My attempt: I noticed the two following blocks:

$\begin{vmatrix}0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\3&0&\ldots&0&0&0&0&\ldots&0&2\end{vmatrix}\;\&\;\begin{vmatrix}2&0&\ldots&0&0&0&0&\ldots&0&3\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\end{vmatrix}$

I switched the blocks because I was dealing with the determinant of the even-order: $$\begin{vmatrix}2&0&\ldots&0&0&0&0&\ldots&0&3\\0&2&\ldots&0&0&0&0&\ldots&3&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\\0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&3&\ldots&0&0&0&0&\ldots&2&0\\3&0&\ldots&0&0&0&0&\ldots&0&2\end{vmatrix}$$ Then I saw we can subtract $j-\text{th column}$ multiplied by $-\frac{3}{2}$ from the $(n-j+1)-\text{column}\;\forall j\in\{1,\ldots,2n\}$

Then I got a $\text{lower-triangular}$ matrix with entries $-\frac{5}{2}$ on the main diagonal.

My final result is: $$D_{2n}=\left(-\frac{5}{2}\right)^{2n}=\left(\frac{5}{2}\right)^{2n}$$ Is this correct?

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    $\begingroup$ I can't tell what your original matrix looks like. If you have dots between a $0$ and a $3$, are there all zeros between? All $3$s? $\endgroup$
    – xxxxxxxxx
    Commented Jan 29, 2020 at 17:39
  • $\begingroup$ @MorganRodgers, we were told this means the rest of the entries are $0$, so I thought it was the practice in the rest of the world as well. $\endgroup$
    – Cherry Blossom
    Commented Jan 29, 2020 at 17:40
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    $\begingroup$ When between two zeros, I would assume this. But for example you have diagonal dots between two 2s. I would assume that means it is all 2s between them. So I am unclear about everything. $\endgroup$
    – xxxxxxxxx
    Commented Jan 29, 2020 at 17:47
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    $\begingroup$ If all the entries of the original matrix are integers, then its determinant should certainly be an integer. $\endgroup$
    – Daniel Schepler
    Commented Jan 29, 2020 at 17:47
  • $\begingroup$ @MorganRodgers, I thought you ask between the entries in one row, yes, the pattern repeats $\endgroup$
    – Cherry Blossom
    Commented Jan 29, 2020 at 17:48

2 Answers 2

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  1. When you switch the two blocks you switch $n$ pairs of rows therefore the determinant comes multiplied by $(-1)^n$.
  2. When you perform the column additions only half of the diagonal entries of the matrix are altered.

The final result should be $(-1)^n2^n(-\frac{5}{2})^n=(-1)^{n+1}2^n(\frac{5}{2})^n=(-1)^{n+1}5^n$

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A recursion can also be helpful here:

  • $n=1$: $D_2 = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = -5$
  • For $n >1$, expanding $D_{2n}$ along the first column gives:

$$D_{2n} = 2\cdot D_{2(n-1)}\cdot 2 - 3 \cdot D_{2(n-1)}\cdot 3 = -5D_{2(n-1)}$$

It follows

$$D_{2n} = (-5)^{n}$$

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