Can we relate these four determinants?

Question

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

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

Answer 1

Define two more determinants \begin{equation} 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}, \end{equation} \begin{equation} 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}. \end{equation} 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 $\,D_4 = 0.\,$

Answer 2

COMMENT.-You can get many relations depending upon the numerical values of $\alpha_i$ and $\beta_i$. For example, noting that $$\begin{equation} D_2 =(-1)(-1)(-1) \det \begin{pmatrix} \alpha_1 &1& \beta_3\\ \alpha_2 &1& \beta_1\\ \alpha_3 &1& \beta_1 \end{pmatrix}=-\det \begin{pmatrix} \alpha_1 &1& \beta_3\\ \alpha_2 &1& \beta_1\\ \alpha_3 &1& \beta_1 \end{pmatrix} \end{equation}$$ 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.

Answer 3

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 $$ \alpha_1+\alpha_2+\alpha_3=2. $$