For $A,I\in \mathbb{R}^{2\times2}$ such that $A^2+A+2I=0$, prove that A is not symmetric

Question

If $A^2+A+2I=0$, where $A,I\in \mathbb{R}^{2\times2}$, I need to show $A$ is not a symmetric matrix.

I have shown that A is a non-singular matrix, using $A^2+A=-2I$. But to show that $A$ isn't symmetric, I'm not getting any ideas.

Answer 1

Assume $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$.

From $A^2+A+2I=0$, we dee that $ad-b^2=\det A=2$ and $a+d=\operatorname{tr}A=-1$. So $ad\ge 2$ and $$ 0\le (a-d)^2=(a+d)^2-4ad\le(- 1)-4\cdot 2<0$$