Otdfbt

If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant

$$beginvmatrix3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2endvmatrix$$

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.

The matrix is the product $AB$, where
$$A=beginpmatrixalpha^2&alpha&1\beta^2&beta&1\
gamma^2&gamma&1endpmatrixmbox and B=beginpmatrix1&1&1\alpha&beta&gamma\
alpha^2&beta^2&gamma^2endpmatrix.$$
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.

Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^2 left( beta-gamma right) ^2
left( beta-alpha right) ^2
$$

Being a $3 times 3$ determinant, it is too small to be attacked with sophisticated techniques, but also too involved to be attacked by brute force. We shall compute it with successive simplifications of its rows and columns. In the following, $R_i$ and $C_j$ will mean the $i$-th row and the $j$-th column, and $[R_i to aR_i + bR_j]$ means to replace the $i$-th row with the linear combination $aR_i + bR_j$ where $a$ and $b$ are numbers. The computation, then, goes as follows:

$$beginalign*
& beginvmatrix
3alpha^2 & beta^2 + alpha beta + alpha^2 & gamma^2 + alpha gamma + alpha^2 \
alpha^2 + alpha beta + beta^2 & 3beta^2 & gamma^2 + beta gamma + beta^2 \
alpha^2 + alpha gamma + gamma^2 & beta^2 + beta gamma + gamma^2 & 3gamma^2endvmatrix = [R_2 to R_2 - R_3] = \[10pt]
& beginvmatrix
3alpha^2 & beta^2 + alpha beta + alpha^2 & gamma^2 + alpha gamma + alpha^2 \
(beta - gamma) (alpha + beta + gamma) & (beta - gamma) (2beta + gamma) & (beta - gamma) (beta + 2gamma) \
alpha^2 + alpha gamma + gamma^2 & beta^2 + beta gamma + gamma^2 & 3gamma^2endvmatrix = \[10pt]
(beta - gamma) & beginvmatrix
3alpha^2 & beta^2 + alpha beta + alpha^2 & gamma^2 + alpha gamma + alpha^2 \
alpha + beta + gamma & 2beta + gamma & beta + 2gamma \
alpha^2 + alpha gamma + gamma^2 & beta^2 + beta gamma + gamma^2 & 3gamma^2endvmatrix = [R_3 to R_3 - R_1] = \[10pt]
(beta - gamma) & beginvmatrix
3alpha^2 & beta^2 + alpha beta + alpha^2 & gamma^2 + alpha gamma + alpha^2 \
alpha + beta + gamma & 2beta + gamma & beta + 2gamma \
(gamma - alpha) (gamma + 2alpha) & (gamma - alpha) (alpha + beta + gamma) & (gamma - alpha) (2gamma + alpha) endvmatrix = \[10pt]
(beta - gamma) (gamma - alpha) & beginvmatrix
3alpha^2 & beta^2 + alpha beta + alpha^2 & gamma^2 + alpha gamma + alpha^2 \
alpha + beta + gamma & 2beta + gamma & beta + 2gamma \
gamma + 2alpha & alpha + beta + gamma & 2gamma + alpha endvmatrix = [C_2 to C_2 - C_1] = \[10pt]
(beta - gamma) (gamma - alpha) & beginvmatrix
3alpha^2 & (beta - alpha) (beta + 2alpha) & gamma^2 + alpha gamma + alpha^2 \
alpha + beta + gamma & beta - alpha & beta + 2gamma \
gamma + 2alpha & beta - alpha & 2gamma + alpha endvmatrix = \[10pt]
(beta - gamma) (gamma - alpha) (beta - alpha) & beginvmatrix
3alpha^2 & beta + 2alpha & gamma^2 + alpha gamma + alpha^2 \
alpha + beta + gamma & 1 & beta + 2gamma \
gamma + 2alpha & 1 & 2gamma + alpha
endvmatrix = [C_3 to C_3 - C_1] \[10pt]
(beta - gamma) (gamma - alpha) (beta - alpha) & beginvmatrix
3alpha^2 & beta + 2alpha & (gamma - alpha) (gamma + 2alpha) \
alpha + beta + gamma & 1 & gamma - alpha \
gamma + 2alpha & 1 & gamma - alpha
endvmatrix = \[10pt]
(beta - gamma) (gamma - alpha)^2 (beta - alpha) & beginvmatrix
3alpha^2 & beta + 2alpha & gamma + 2alpha \
alpha + beta + gamma & 1 & 1 \
gamma + 2alpha & 1 & 1
endvmatrix = [C_3 to C_3 - C_2] \[10pt]
(beta - gamma) (gamma - alpha)^2 (beta - alpha) & beginvmatrix
3alpha^2 & beta + 2alpha & gamma - beta \
alpha + beta + gamma & 1 & 0 \
gamma + 2alpha & 1 & 0
endvmatrix = \[10pt]
- (beta - gamma)^2 (gamma - alpha)^2 (beta - alpha) & beginvmatrix
3alpha^2 & beta + 2alpha & 1 \
alpha + beta + gamma & 1 & 0 \
gamma + 2alpha & 1 & 0
endvmatrix = \[10pt]
- (beta - gamma)^2 (gamma - alpha)^2 (beta - alpha)^2
endalign*
$$

where the last $3 times 3$ determinant has been expanded along the $3$rd column.

Since the determinant is the negative of the square of a product of real numbers, its maximum will be $0$, reached whenever any two of $alpha, beta, gamma in [-3,10]$ are equal.

Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.