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 ${\left(P^{-1}AP-I_3\right)}^2$ is $\alpha {\omega }^2$, then the value of $\alpha$ is equal to ______.

Answer 1

Answer: 36

Let $M={\left(P^{-1}AP-I\right)}^2$
$={\left(P^{-1}AP\right)}^2-2P^{-1}AP+I$
$=P^{-1}{A}^{2}P-2P^{-1}AP+I$
$PM=A^{2}P-2AP+P$
$=\left(A^2-2A\cdot I+I^2\right)P$
$\Rightarrow \mathrm{Det}(PM)=\mathrm{Det}\left((A-I{)}^2\times P\right)$
$\Rightarrow$ DetP.DetM $=\mathrm{Det}(A-I{)}^2\times \mathrm{Det}(P)$
$\Rightarrow$ Det $M=(\mathrm{Det}(A-I){)}^2$
Now $A-I=\left[\begin{array}{ccc}1& 7& w^2\\ -1& -w-1& 1\\ 0& -w& -w\end{array}\right]$
$\mathrm{Det}(A-I)=\left(w^2+w+w\right)+7(-w)+w^3=-6u$
$\mathrm{Det}((A-I){)}^2=36w^2$
$\Rightarrow \alpha =36$