# Prove the following

Question:

LetÂ

$P=\left[\begin{array}{ccc}-30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14\end{array}\right]$ 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 $\mathrm{I}_{3}$ be the identity

matrix of order 3 . If the determinant of the matrix $\left(\mathrm{P}^{-1} \mathrm{AP}-\mathrm{I}_{3}\right)^{2}$ is $\alpha \omega^{2}$, then the value of $\alpha$ is equal to________.

Solution:

Let $\mathrm{M}=\left(\mathrm{P}^{-1} \mathrm{AP}-\mathrm{I}\right)^{2}$

$=\left(\mathrm{P}^{-1} \mathrm{AP}\right)^{2}-2 \mathrm{P}^{-1} \mathrm{AP}+\mathrm{I}$

$=\mathrm{P}-1 \mathrm{~A}^{2} \mathrm{P}-2 \mathrm{P}^{-1} \mathrm{AP}+\mathrm{I}$

$\mathrm{PM}=\mathrm{A}^{2} \mathrm{P}-2 \mathrm{AP}+\mathrm{P}$

$=\left(\mathrm{A}^{2}-2 \mathrm{~A} \cdot \mathrm{I}+\mathrm{I}^{2}\right) \mathrm{P}$

$\Rightarrow \operatorname{Det}(P M)=\operatorname{Det}\left((A-I)^{2} \times P\right)$

$\Rightarrow \operatorname{DetP.DetM}=\operatorname{Det}(A-I)^{2} \times \operatorname{Det}(P)$

$\Rightarrow \operatorname{Det} M=(\operatorname{Det}(A-I))^{2}$

Now $A-I=\left[\begin{array}{ccc}1 & 7 & w^{2} \\ -1 & -w-1 & 1 \\ 0 & -w & -w\end{array}\right]$

$\operatorname{Det}(A-I)=\left(w^{2}+w+w\right)+7(-w)+w^{3}=-6 w$

$\operatorname{Det}((A-I))^{2}=36 w^{2}$

$\Rightarrow \alpha=36$