We know that $A$ is a $3\times3$ matrix on $R$ such that $A^{2018}=O_3$. We also know there exists a matrix $B$ ($3\times3$ on $R$) such that $A^{2017}B+BA=I_3$
Which one of these is true?
a) $A=A^\top$
b) $\operatorname{tr}(A)=0$
c) $A^2=O_3$
d) $A^3=O_3$
e) $A^{1009}=O_3$
f) there is no matrix $A$ with these properties.
I noticed that $\det A$ has to be zero (from $A^{2018}=O_3$). Multiplying $A^{2017}B+BA=I_3$ with $A$ to the left gives us $ABA=A$. I have no idea what to do from here. Thanks for your help!
$A$ does not exist.
Using $A^{2018}=0$, if we left-multiply both sides of $A^{2017}B+BA=I$ by $A$, we obtain $ABA=A$. Therefore $BABA=BA$ and in turn, $(I-BA)^2=I-BA$. But then $$ I-BA=(I-BA)^2=(A^{2017}B)(A^{2017}B)=A^{2016}(ABA)A^{2016}B=A^{2016}AA^{2016}B=0. $$ Hence $BA=I$. Now $A$ is both singular (as $A^{2018}=0$) and nonsingular (as $BA=I$). This is impossible. Thus $A$ does not exist and option (f) is true. In turn, the other five options are also true, because they are vacuous truths.
Remark: a university-level solution. Since $A^{2018}=0$, $A$ is nilpotent. Thus $A^3=0$ (because $A$ is $3\times3$) and the equation $A^{2017}B+BA=I$ implies that $BA=I$. But then $A$ both singular (as it is nilpotent) and nonsingular (as $BA=I$), which is impossible. Hence $A$ does not exist in the first place.