Question

Let $P=\left[\begin{array}{ccc}-30& 20& 56\\ 90& 140& 112\\ 120& 60& 14\end{array}\right]$ and $A=\left[\begin{array}{ccc}2& 7& {\omega }^2\\ -1& -\omega & 1\\ 0& -\omega & -\omega +1\end{array}\right]$ where $\omega =\frac{-1+i\sqrt{3}}{2}$ and $I_3$ be the identity matrix of order 3. If the determinant of the matrix $(P^{-1}AP-I_3{)}^2$ is $\alpha {\omega }^2$, then the value of $\alpha$ is equal to

Answer 1

$|P^{-1}AP-1{|}^2=|(P^{-1}AP-1)(P^{-1}AP-1)|=|P^{-1}AP{P}^{-1}AP-2P^{-1}AP+1|=|P^{-1}{A}^{2}P-2P^{-1}AP+P^{-1}IP|=|P^{-1}(A^2-2A+1)P|=|P^{-1}(A-I{)}^{2}P|=|P^{-1}||A-1{|}^2|P|=|A-I{|}^2$
$={\left|\begin{array}{ccc}1& 7& {\omega }^2\\ -1& -\omega -1& 1\\ 0& -\omega & -\omega \end{array}\right|}^2$
$=\left(1\right(\omega (\omega +1)+\omega )-7\omega +{\omega }^2\omega {)}^2=({\omega }^2+2\omega -7\omega +1{)}^2=({\omega }^2-5\omega +1{)}^2=(-6\omega {)}^2=36{\omega }^2\Rightarrow \alpha =36$