If $A$ is a matrix such that $A^{2}+A+2I=O$, then $A$ can't be skew symmetric. (True/false)
When $A$ is odd order matrix then the statement is true, since $A$ is non singular. $( |A||A+I| = (-2)^{n})$ and skew symmetric matrix of odd order is singular.
How to check for even case$?$
Best Answer
Let $A$ be a matrix such that $$A^2+A+2I=0\tag{1}.$$ Then also $$0=0^{\top}=(A^2+A+2I)^{\top}=(A^{\top})^2+A^{\top}+2I.$$ Suppose $A$ is skew-symmetric, so that $A^{\top}=-A$. Then it follows that $$0=(-A)^2+(-A)+2I=A^2-A+2I,\tag{2}$$ and subtracting this from $(1)$ then yields $2A=0$ and hence $A=0$. But clearly this does not satisfy $(1)$, a contradiction. So $A$ can't be skew symmetric.