Let P=pmatrix-30&20&5690&140&112120&60&14pmatrix and A=pmatrix2&7&^2-1&-&10&-&-+1pmatrix, …

Question

Let $P=\begin{pmatrix}-30&20&56\90&140&112\120&60&14\end{pmatrix}$ and $A=\begin{pmatrix}2&7&\omega^2\-1&-\omega&1\0&-\omega&-\omega+1\end{pmatrix}$, where $\omega=\dfrac{-1+i\sqrt3}2$, and $I_3$ the identity matrix of order 3. If the determinant of the matrix $(P^{-1}AP-I_3)^2$ is $\alpha\omega^2$, then $\alpha$ equals.........

Answer: 36

Accepted Answer

$$\det\big((P^{-1}AP-I_3)^2\big)=\det(P^{-1}AP-I_3)^2 = \big|P^{-1}(A-I_3)P\big|^2 = |A-I_3|^2$$
(since $|P^{-1}||P|=1$).

$A-I_3=\begin{pmatrix}1&7&\omega^2\-1&-\omega-1&1\0&-\omega&-\omega\end{pmatrix}$. Expanding (using $\omega^2+\omega+1=0$), the determinant works out to $-6\omega$, so
$$|A-I_3|^2=36\omega^2=\alpha\omega^2 \Rightarrow \alpha=36.$$

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