# A problem about determinant and matrix

Suppose $$a_{0},a_{1},a_{2}\in\mathbb{Q}$$, such that the following determinant is zero, i.e.

$$\left |\begin{array}{cccc}\\ a_{0} &a_{1} & a_{2} \\ \\ a_{2} &a_{0}+a_{1} & a_{1}+a_{2} \\ \\ a_{1} &a_{2} & a_{0}+a_{1}\\ \end{array}\right| =0$$

Problem. Show that $$a_{0}=a_{1}=a_{2}=0.$$

I think it's equivalent to show that the rank of the matrix is $$0$$, and it's easy to show the rank cannot be $$1$$. But I have no idea how to show that the case of rank 2 is impossible. So, is there any better idea?

• This is false. Try $a_1= a_0+a_2=0$.
– abx
Feb 28, 2021 at 5:18
• @abx Substituting $a_1=0$ and $a_2=-a_0$ gives determinant $-a_0^3$. So that isn't a counterexample. Feb 28, 2021 at 5:22
• @Brendan McKay: I find $a_0^3-a_0a_2^2$.
– abx
Feb 28, 2021 at 5:29
• @abx With two substitutions only one variable should be left. I'm using Maple. Feb 28, 2021 at 5:32
• There is no counterexample with any of the variables equal to 0. For example if $a_0=0$ and $a_1=ca_2$ then the determinant is $a_2^3(c^3-c^2+1)$. The cubic is irreducible so only $a_2=0$ makes this 0. There are 6 cases like this. Feb 28, 2021 at 5:36

The determinant $$a_0^3+2a_0^2a1+a_0a_1^2-3a_0a_1a_2-a_0a_2^2+a_1^3-a_1^2a_2+a_2^3$$ is odd unless $$a_0,a_1,a_2$$ are all even.
This contradiction shows that $$a_0=a_1=a_2=0$$ is only rational solution.
Incidentally, the result does not hold modulo an arbitrary prime. For example $$a_0=1, a_1=2, a_2=3$$ works mod 5.
You can check that the determinant is the product of three terms $$a_2+ma_1+a_0/m$$ as $$m$$ runs over the three roots of the cubic $$m^3-m-1$$, of which one is real and the other two are complex conjugates. This does not immediately answer the question (which Brendan McKay has done anyway) but it may be useful context.