Let $P=\begin{bmatrix} -30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14 \end{bmatrix}$ and…

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 $\omega=\frac{-1+i\sqrt{3}}{2}$, and $I_3$ be the identity matrix of order $3$. If the determinant of the matrix $\left((P^{-1}AP)-I_3\right)^2$ is $\alpha\omega^2$, then the value of $\alpha$ is equal to _________.

Solution

$M=(P^{-1}AP-I)^2$ $=(P^{-1}AP)^2-2P^{-1}AP+I$ $=P^{-1}A^2P-2P^{-1}AP+I$ $\therefore (P^{-1}AP)^2=(P^{-1}AP)(P^{-1}AP)$ Multiplying with $P$ on both sides $\Rightarrow PM=A^2P-2AP+P$ $=(A^2-2A+I)P$ $\Rightarrow \text{Det}(PM)=\text{Det}((A-I)^2P)$ $\Rightarrow \text{Det}(PM)=\text{Det}((A-I)^2)\text{Det}(P)$ $\Rightarrow \text{Det}(M)=\text{Det}((A-I)^2)^2$ Now $A-I=\begin{bmatrix} 1 & 7 & w^2 \\ -1 & -w-1 & 1 \\ 0 & -w & -w \end{bmatrix}$ $\text{Det}(A-I)=w^2+w+w+7(-w)+w^3=-6w$ $\text{Det}((A-I)^2)=36w^2$ $\Rightarrow \alpha=36$

Asked in: JEE Main 2021 (16 Mar Shift 1)

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