Can we relate these four determinants?

$$$$D_1 =\det \begin{pmatrix} \alpha_1 &1& \beta_2\\ \alpha_2 &1& \beta_3\\ \alpha_3 &1& \beta_1 \end{pmatrix}$$$$ $$$$D_2 = \det \begin{pmatrix} \beta_1 &1& \alpha_2\\ \beta_2 &1& \alpha_3\\ \beta_3 &1& \alpha_1 \end{pmatrix},$$$$ $$$$D_3 = \det \begin{pmatrix} \alpha_1 & \alpha_3 & \beta_2\\ \alpha_2 & \alpha_1 & \beta_3\\ \alpha_3 & \alpha_2 & \beta_1 \end{pmatrix},$$$$ $$$$D_4 = \det \begin{pmatrix} \alpha_1 & \beta_3 & \beta_2\\ \alpha_2 & \beta_1 & \beta_3\\ \alpha_3 & \beta_2 & \beta_1 \end{pmatrix},$$$$

Note that $$\alpha_i, \beta_i$$ are real numbers.

Can we relate and connect the four determinants $$D_i$$, for $$i=1,2,3,4$$? Can we point out some relations between them?

For example: If $$D_1=D_2=0$$, then it implies that $$D_3=0$$ or $$D_4=0$$.[this property is already known] I am searching for other connections and relationship between D_i.

Something like this but not limited to $$D_1 = 0, D_2 \neq 0$$, $$D_2 \neq 0, D_1 =0$$, $$D_1 >0, D_2 = 0$$ what happens to $$D_3, D_4$$?

Any interesting properties connecting the four determinants are welcome!

EDIT

Mainly,if we fix $$D_1<0$$ and $$D_2<0$$ then what happens to $$D_3, D_4$$ ? How they behave?

The question has arised from a question on MathOverflow: https://mathoverflow.net/questions/426170/is-it-impossible-for-determinants-of-these-matrices-to-both-be-negative

• Nice, it would be interesting to see those relations actually! Commented Nov 12, 2022 at 12:17
• Why is this of interest? Commented Nov 12, 2022 at 16:38
• Please see edited question Commented Nov 13, 2022 at 0:32

Indeed, if $$D_1=D_2=0$$, then either $$D_3=0$$ or $$D_4=0$$.

To see this, consider two cases.

Case 1: $$\alpha_3=\alpha_1$$. Then $$D_1=0$$ says $$(\alpha_1-\alpha_2)(\beta_1-\beta_2)=0$$. If $$\alpha_1=\alpha_2$$, then $$D_3=0$$. If $$\beta_1=\beta_2$$, then $$D_2=0$$ implies $$D_3=0$$ or $$D_4=0$$.

Case 2: $$\alpha_3-\alpha_1\neq 0$$. Then $$D_1=0$$ says $$\beta_3:= -\frac{1}{\alpha_3- \alpha_1}(\alpha_1\beta_1 - \alpha_2\beta_1 + \alpha_2\beta_2 - \alpha_3\beta_2)$$

Eliminating $$\beta_2$$ from $$D_2=0$$ we obtain $$D_4=0$$.

Edit: Here is a new one. If $$D_1=D_2=D_3=1$$, or if $$D_1=D_2=D_3=-1$$, then $$D_4=0$$ if and only if $$\alpha_1+\alpha_2+\alpha_3=2.$$

• Thanks! but this property is already known] I am searching for other connections and relationship between D_i Commented Nov 12, 2022 at 16:33
• OK, then here is a new one. If $D_1=0$, $D_2=D_3=1$, then $D_4\neq 0$. Commented Nov 12, 2022 at 17:14
• Thanks, please see. the edited answer. Commented Nov 13, 2022 at 0:38
• Your question "if we fix $D_1<0,D_2<0$, then what happens to $D_3,D_4$ " is too vague. Can you be more precise? For example, saying $D_1=D_2=-1$ has a lot of consequences for the relation of $D_3$ and $D_4$. But which consequences do you want? Commented Nov 13, 2022 at 9:43

COMMENT.-You can get many relations depending upon the numerical values of $$\alpha_i$$ and $$\beta_i$$. For example, noting that $$$$D_2 =(-1)(-1)(-1) \det \begin{pmatrix} \alpha_1 &1& \beta_3\\ \alpha_2 &1& \beta_1\\ \alpha_3 &1& \beta_1 \end{pmatrix}=-\det \begin{pmatrix} \alpha_1 &1& \beta_3\\ \alpha_2 &1& \beta_1\\ \alpha_3 &1& \beta_1 \end{pmatrix}$$$$ and developping with the first row, you have $$D_1-D_2+D_3+D_4=\alpha_1X+\alpha_2Y+\alpha_3Z=N$$ where $$X=\beta_1-\beta_2+\beta_2-\beta_1+\beta_3-\beta_2=\beta_3-\beta_2$$, etc.

All being function of reals involved. However maybe there are relations (like what happen with $$X$$) of interest with just the literal values.

• Thanks, please see. the edited answer. Commented Nov 13, 2022 at 0:39

Define two more determinants $$$$D_5 := \det \begin{pmatrix} \alpha_1 & \alpha_3 & \beta_1\\ \alpha_2 & \alpha_1 & \beta_2\\ \alpha_3 & \alpha_2 & \beta_3 \end{pmatrix},$$$$ $$$$D_6 := \det \begin{pmatrix} \alpha_3 & \beta_3 & \beta_2\\ \alpha_1 & \beta_1 & \beta_3\\ \alpha_2 & \beta_2 & \beta_1 \end{pmatrix}.$$$$ Define two more values $$A := \alpha_1^2 + \alpha_2^2 + \alpha_3^2 - \alpha_1\alpha_2 - \alpha_1\alpha_3 - \alpha_2\alpha_3, \\ B := \beta_1^2 + \beta_2^2 + \beta_3^2 - \beta_1\beta_2 - \beta_1\beta_3 - \beta_2\beta_3.$$ Verify that $$B\,D_3 = D_1D_6 + D_2D_4, \quad D_3 = D_1\alpha_1 + D_2\alpha_2 + \beta_3A = \\ D_1\alpha_2 + D_2\alpha_3 + \beta_1A = D_1\alpha_3 + D_2\alpha_1 + \beta_2A.$$ Suppose that $$\,D_1 = D_2 = 0.\,$$ The first case is if $$\,\beta_1 = \beta_2 = \beta_3 = 0,\,$$ then $$\,D_3 = 0.\,$$ Second case is if $$\,A = 0\,$$ then $$\,D_3 = 0.\,$$ Third case is if $$\,A \ne 0.\,$$ Verify that $$\, A B = D_1^2 - D_1D_2 + D_2^2 \,$$ which implies that $$\,B=0\,$$ but the identity with $$\,D_6\,$$ implies that $$\,D_3 = 0\,$$ again.

Thus, in all three cases, $$\,D_1 = D_2 = 0\,$$ implies that $$\,D_3 = 0.\,$$

Similar identities hold for $$\,D_4 = 0.\,$$ Thus, if $$\,D_1 = D_2 = 0\,$$ then $$\,D_3 = 0\,$$ or $$\,D_4 = 0.\,$$

• I have a counterexample. Let $\alpha_1=1,\alpha_2=2,\alpha_3=3$ and $\beta_i=1$ for $i=1,2,3$. Then $D_1=D_2=D_4=0$, but $D_3=3$. But one of $D_3$ and $D_4$ has to be zero. Commented Nov 12, 2022 at 16:20
• @DietrichBurde Thanks for that comment! I will have to fix my logic. Commented Nov 12, 2022 at 16:25
• You still say "In all cases, $D_1=D_2=0$ implies that $D_3=0$". But this is not true, right? Commented Nov 12, 2022 at 16:30
• @BAYMAX Yes, but I have other relations. For example, $BD_3=D_1D_6 +D_2D_4.$ Commented Nov 12, 2022 at 16:36
• @DietrichBurde Do you have a counter-example? Commented Nov 12, 2022 at 16:37