TITLE: Largest value of determinant
QUESTION [5 upvotes]: If $\alpha,\beta,\gamma \in [-3,10].$ Then largest value of the determinant
$$\begin{vmatrix}3\alpha^2&\beta^2+\alpha\beta+\alpha^2&\gamma^2+\alpha\gamma+\alpha^2\\\\
\alpha^2+\alpha\beta+\beta^2& 3\beta^2&\gamma^2+\beta\gamma+\beta^2\\\\
\alpha^2+\alpha\gamma+\gamma^2& \beta^2+\beta\gamma+\gamma^2&3\gamma^2\end{vmatrix}$$

Try: I am trying to break that determinant into product of 2 determinants but not able to break it.
Could someone help me in this question? Thanks.

REPLY [7 votes]: The matrix is the product $AB$, where
$$A=\begin{pmatrix}\alpha^2&\alpha&1\\\beta^2&\beta&1\\
\gamma^2&\gamma&1\end{pmatrix}\mbox{ and }B=\begin{pmatrix}1&1&1\\\alpha&\beta&\gamma\\
\alpha^2&\beta^2&\gamma^2\end{pmatrix}.$$
So its determinant is the product of $\det(A)$ and $\det(B)$.
The matrix $A$ can be obtained from $B$ by exchanging the first and the third row and then transposing the result. Therefore $\det(A)=-\det(B)$ and the determinant
of the given matrix is $-\det(B)^2$. This is a non-positive expression and its
maximal value $0$ is attained if and only if $\det(B)=0$.
Now $B$ is a Vandermonde matrix and its determinant is known to be
$(\gamma-\beta)(\gamma-\alpha)(\beta-\alpha)$. This determinant is zero and thus the given determinant attains its maximal value if and only if two or more of the numbers $\alpha,\beta,\gamma$ are equal.