Question

Let $A$ be a square matrix of order $3$ such that $A+A^{\top }=$
$\left[\begin{array}{ccc}10& 4& 6\\ a_{21}+a_{12}& 6& a_{23}+a_{32}\\ a_{31}+a_{13}& 8& 4\end{array}\right]$,
where $a_{12},a_{23},a_{31}$ are positive roots of the equation $x^3-6x^2+px-8=0,\forall p\in R$, then the absolute value of $|A|$ is equal to

Answer 1

Answer: 37

Clearly, $A+A^T$ is a symmetric matrix
Also, $a_{12}+a_{23}+a_{31}=6,a_{12}{a}_{23}{a}_{31}=8$
$AM=GM\Rightarrow a_{12}=a_{23}=a_{31}=2$
$a_{21}+a_{12}=4\Rightarrow a_{21}+a_{12}=2$
$a_{31}+a_{13}=6\Rightarrow a_{31}+a_{13}=3$
$a_{23}+a_{32}=8\Rightarrow a_{23}+a_{32}=4$
For diagonal elements
$2a_{11}=10\Rightarrow a_{11}=5$
$2a_{22}=6\Rightarrow a_{22}=3$
$2a_{33}=4\Rightarrow a_{33}=2$
$A=\left[\begin{array}{ccc}5& 2& 3\\ 2& 3& 4\\ 3& 4& 2\end{array}\right]$
$|A|=5(6-16)-2(4-12)+3(8-9)$
$=-50+16-3$
$=-37$