Let $A=[a_{ij}]_{3\times 3}$ be a matrix.If $A+A^T= \begin{bmatrix} 6 & 4 & 4\\ a_{21}+a_{12} & 10 & a_{23}+a_{32}\\ a_{31}+a_{13} &4&8 \end{bmatrix}$ where $a_{12},a_{23}$ and $a_{31}$ are the positive roots of the equation $x^3-6x^2+px-8=0,p\in R$ then find the value of $\det(A)$.Here $A^T$ denotes the transpose of matrix $A.$
My Attempt:Let $A=\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} &a_{32}&a_{33}
\end{bmatrix}$,then $A^T=\begin{bmatrix}
a_{11} & a_{21} & a_{31}\\
a_{12} & a_{22} & a_{32}\\
a_{13} &a_{23}&a_{33}
\end{bmatrix}$
$A+A^T=\begin{bmatrix}
2a_{11} & a_{12}+a_{21} & a_{13}+a_{31}\\
a_{12}+a_{21} & 2a_{22} & a_{23}+a_{32}\\
a_{13}+a_{31} &a_{23}+a_{32}&2a_{33}
\end{bmatrix}$
but $A+A^T=
\begin{bmatrix}
6 & 4 & 4\\
a_{21}+a_{12} & 10 & a_{23}+a_{32}\\
a_{31}+a_{13} &4&8
\end{bmatrix}$
so $a_{11}=3,a_{22}=5,a_{33}=4,a_{21}+a_{12}=4,a_{31}+a_{13}=4,a_{23}+a_{32}=4$
Also $a_{12},a_{23} \text{and} a_{31}$ are the positive roots of the equation $x^3-6x^2+px-8=0$,so
$a_{12}+a_{23}+a_{31}=6,a_{12}.a_{23}.a_{31}=8,a_{12}a_{23}+a_{23}a_{31}+a_{31}a_{12}=p$
For finding the determinant of $A$,i need to find the values of $a_{12},a_{13},a_{21},a_{23},a_{31},a_{32}$ also which i am not able to find.
Please help me.Thanks.
Since $a_{12},a_{23},a_{31}$ are supposed to be real and positive we have, from the AM-GM inequality $$\frac{a_{12}+a_{23}+a_{31}}{3}\ge\sqrt[3]{a_{12}\cdot a_{23}\cdot a_{31}}$$ last inequality becomes an equality iif $a_{12}=a_{23}=a_{31}$.
Now, you have found that $a_{12}+a_{23}+a_{31}=6$ and $a_{12}\cdot a_{23}\cdot a_{31}=8$, which means $$\frac{a_{12}+a_{23}+a_{31}}{3}=2=\sqrt[3]{a_{12}\cdot a_{23}\cdot a_{31}}$$ Then, $$a_{12}=a_{23}=a_{31}=2$$ I leave to you the remaining part.