JEE Main 2021 (Online) 16th March Morning ShiftQuestion 1
Let $P=\begin{bmatrix}-30&20&56\\90&140&112\\120&60&14\end{bmatrix}$ and $A=\begin{bmatrix}2&7&\omega^{2}\\-1&-\omega&1\\0&-\omega&-\omega+1\end{bmatrix}$ where $\omega=\dfrac{-1+i\sqrt{3}}{2}$, 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 $\_\_\_\_\_\_\_\_$ .
Matrices
Answer$36$
Solution
$\det\left((P^{-1}AP-I)^2\right)=\left(\det(A-I)\right)^2$. Expanding, $\det(A-I)=3-3\sqrt3\,i=-6\omega$, so the determinant is $(-6\omega)^2=36\omega^2$ and $\alpha=36$.