Let $P=\left[\begin{array}{ccc}-30 & 20 & 56 \ 90 & 140 & 112 \ 120 & 60 & 14\end{array} ight]$ a

Question

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

Professor PC Thomas Classes · Answer

(36) : We have, $\left|\left(P^{-1} A P-I_{3}ight)^{2}ight|$\ $=\left|P^{-1} A P-I_{3}ight|^{2}$\ $\Rightarrow \quad \alpha \omega^{2}=\left|P^{-1} A P-P^{-1} Pight|^{2} \quad\left(\because I=P^{-1} Pight)$\ $=\left|P^{-1}(A P-P)ight|^{2}=\left|P^{-1}ight|^{2}|A P-P|^{2}$\ $=\left|P^{2}ight|^{-1}|A-I| 2|P|^{2}=\frac{1}{|P|^{2}}|A-I|^{2}|P|^{2}$ $$\begin{equation*} \Rightarrow \quad \alpha \omega^{2}=|A-I|^{2} 	ag{i} \end{equation*}$$ Now, $|A-I|=\left|\begin{array}{ccc}1 & 7 & \omega^{2} \ -1 & -\omega-1 & 1 \ 0 & -\omega & -\omega\end{array}ight|=-6 \omega$\ $\left(\because \quad 1+\omega+\omega^{2}=0ight)$\ $\Rightarrow \quad|A-I|^{2}=36 \omega^{2}$\ $\Rightarrow \quad \alpha \omega^{2}=36 \omega^{2}$\ (Using (i))\ $\Rightarrow \quad \alpha=36$\