# Let

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 $\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2}$, 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 $\mathrm{M}=\left({ }^{\mathrm{P}^{-1}} \mathrm{AP}-\mathrm{I}\right)^{2}$

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \quad \alpha=36$