If $A\in M_3\left(\mathbb C\right)$ is an invertible matrix such that $2A^2=4A+A^3$ If $A\in M_3\left(\mathbb C\right)$ is an invertible matrix such that
 $$2A^2=4A+A^3$$
Then which of the following is(are)correct:
(A) $\det (A)=8;$
(B) $det\left(adj\left(\frac A2\right)\right)=1;$
(C) $tr\left(A-2I_3\right)^3=24;$
(D) $adj(A)=A^2$
My working:
From given information we get $A(A^2-2A+4I_3)=O_3$
$\implies A^2-2A+4I_3=O_3$
$\implies (A+2\omega I_3)(A+2\omega^2 I_3)=O_3$
$\implies$ Possible Eigen values of $A$ can be $-2\omega, -2\omega^2$
Now what to do

From the given information, $A(A^2-2A+4I_3)=O_3$
$\implies A^2-2A+4I_3=O_3$
$\implies (A+2\omega I_3)(A+2\omega^2 I_3)=O_3$
$\implies$ eigen values of $A$ can be $-2\omega,-2\omega,-2\omega^2$ or $-2\omega,-2\omega^2,-2\omega^2.$ Where $\omega$ is a non real cube root of unity.
$\implies \det A=$ Product of eigen values of $A$ = $(-2\omega)\times(-2\omega)\times(-2\omega^2)$ or $(-2\omega)\times(-2\omega^2)\times(-2\omega^2)=-8\omega$ or $-8\omega^2.$
$\implies$ Option (A) is incorrect.
As in previous solution we have $A^3=-8I_3$
$\implies A^2=-8A^{-1}=-8\frac{1}{\det A}adj(A)=\omega^2 adj(A)$ or $\omega adj(A)\implies adj(A)=\omega A^2$ or $\omega^2 A^2$
$\implies$ option (D) is incorrect.
Now $adj\left(\frac A2\right)=\left(\frac 12\right)^2 adj(A)=\frac 14 \omega A^2$  or  $\frac 14 \omega^2 A^2$ 
$\implies\det \left(adj\left(\frac A2\right)\right)=\left(\frac 14 \omega\right)^3 (\det A)^2$  or  $\left(\frac 14 \omega^2\right)^3 (\det A)^2=\omega^2$  or $\omega$
$\implies$ Option (B) is incorrect.
So only correct option is (C) as demonstrated in previous solution.
