# Computation of the $2n$-th order determinant

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?

• 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? Commented Jan 29, 2020 at 17:39
• @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. Commented Jan 29, 2020 at 17:40
• 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. Commented Jan 29, 2020 at 17:47
• If all the entries of the original matrix are integers, then its determinant should certainly be an integer. Commented Jan 29, 2020 at 17:47
• @MorganRodgers, I thought you ask between the entries in one row, yes, the pattern repeats Commented Jan 29, 2020 at 17:48

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 $$(-1)^n2^n(-\frac{5}{2})^n=(-1)^{n+1}2^n(\frac{5}{2})^n=(-1)^{n+1}5^n$$

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

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