Show that the determinant of the matrix \begin{pmatrix} a && -c && -b \\ b && a - 2c && -c -2b \\ c && b && a -2c \end{pmatrix} is non zero for all integers $a,b,c$ where $abc \ne 0$

There is an interesting way to do this by using integral domains. It is easy to see that the polynomial $t^3 + 2t + 1$ is prime in $\mathbb{Z}[t]$ so the ring $\mathbb{Z}[t]/\langle t^3+2t+1\rangle$ is an integral domain and isomorphic to the ring $\{a + bu + cu^2 \vert \text{$a,b,c$ are integers}\}$ where $u$ is a root of the equation $t^3 +2t + 1 =0$.

Now consider integers $a,b,c$ and $x,y,z$ such that $abc \ne 0$ and $(a +bu + cu^2)(x + yu + zu^2) = 0$. These are element of an integral domain so we must have that $x = y = z = 0$ since at least one of $a,b$ and $c$ are nonzero. But expanding the above equation we get a system of linear equations in terms of $x,y$ and $z$.

$$ \begin{aligned} ax- cy -bz &= 0 \\ bx+ (a-2c)y + (-c-2b)z &= 0\\ c x+b y+ (a-2c)z &= 0 \end{aligned} $$

If the determinant is $0$ then this equation will have a nontrivial solution which is not possible.

This strategy requires the use of integral domains which is an abstract tool.

Are there any elementary ways to solve this? It gets messy when we try to do row operations or try to expand the determinant.

Update : After Carl Schildkraut's and JimmyK4542's answer.

A curious observation:

We can also prove that the element $t^3 + rt + 1$ is prime in $\mathbb{Z}[t]$ for $r \ne 0, -2$. Therefore by similar arguments we can say that the determinant of the matrix \begin{pmatrix} a && -c && -b \\ b && a - rc && -c -rb \\ c && b && a -rc \end{pmatrix} is non zero for all integers $a,b,c,r$ where $abc \ne 0$ and $r \ne 0,-2$.

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    $\begingroup$ The determinant is $a^3-b^3+c^3+2ab^2-2bc^2-4ca^2+4c^2a+3abc$. $\endgroup$
    – user1551
    Jun 26, 2021 at 9:33
  • $\begingroup$ @ user1551 Can we use this expression in anyway ? (for example factoring or completing squares) $\endgroup$ Jun 26, 2021 at 10:07
  • $\begingroup$ I made another observation also, $t^3 + mt + 1$ is prime in $\mathbb{Z}[t]$ for $m \ne 0,-2$, so working with this gives more general matrix. $\endgroup$ Jun 26, 2021 at 10:09
  • $\begingroup$ I have no idea, but number theorists may see something in it. An answer below considers congruence modulo $3$, for instance. $\endgroup$
    – user1551
    Jun 26, 2021 at 11:38

2 Answers 2


This might be equivalent to what you did, but I'll go ahead and write this as an answer anyway.

Let $X = \begin{bmatrix}0 & 0 & -1 \\ 1 & 0 & -2 \\ 0 & 1 & 0\end{bmatrix}$. Then, $X^2 = \begin{bmatrix}0 & -1 & 0 \\ 0 & -2 & -1 \\ 1 & 0 & -2\end{bmatrix}$, and so, $$Y := \begin{bmatrix}a & -c & -b \\ b & a - 2c & -c -2b \\ c & b & a -2c\end{bmatrix} = aI+bX+cX^2.$$

The eigenvalues of $X$ are the three roots $\lambda_1,\lambda_2,\lambda_3$ of $\lambda^3+2\lambda+1 = 0$.

Since $Y = aI+bX+cX^2$, the eigenvalues of $Y$ are $a+b\lambda_k+c\lambda_k^2$ for $k = 1,2,3$.

Now, suppose $\det(Y) = 0$ for some integers $a,b,c$ with $abc \neq 0$. Then, $0$ must be an eigenvalue of $Y$, and so, $a+b\lambda_k+c\lambda_k^2 = 0$ for some $k \in \{1,2,3\}$. But by using the quadratic formula, we have $\lambda_k = \dfrac{-b \pm \sqrt{b^2-4ca}}{2c}$.

So now you just need to prove that none of the three roots of $\lambda^3+2\lambda+1 = 0$ can be expressed in the form $\dfrac{-b \pm \sqrt{b^2-4ca}}{2c}$ for non-zero integers $a,b,c$. This just requires showing that $\lambda^3+2\lambda+1$ is irreducible in $\mathbb{Z}[\lambda]$, which only requires the rational root theorem and checking that $\pm 1$ are not roots.

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    $\begingroup$ Perhaps I am missing something, but it seems easier to use the rational root theorem to prove that $p(\lambda)=\lambda^3+2\lambda+1$ is irreducible over $\mathbb Q$, and conclude that $p$ is the minimal polynomial of $X$ over $\mathbb Q$. $\endgroup$
    – user1551
    Jun 26, 2021 at 8:57

Say we have a counterexample. If $3$ divides each of $a,b,c$, then we can divide each by $3$ without consequence, and obtain another counterexample. We may do this until $3$ fails to divide one of $a,b,c$.

However, we may calculate that $3$ fails to divide the determinant in any other case. To do this we would ordinarily have $26$ cases. However, since our polynomial is homogeneous in $a,b,c$, we only need to check triples up to scaling, which halves the number of cases. When two of $\{a,b,c\}$ are $0\bmod 3$ this check is easy, as only a single nonzero term persists in the determinant expansion. This reduces it to only $10$ cases, each of which can be checked by hand (it may even be possible to reduce further without much computation).

Remark. This strategy works for any prime $p$ for which $t^3+2t+1$ is irreducible in $\mathbb F_p[t]$. This isn't true for $p=2$, so $p=3$ yields the fewest number of cases.


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