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 ____________.
Option: 1 -6
Option: 2 36
Option: 3 -36
Option: 4 6

Answers (1)

S Suraj Bhandari

$\begin{aligned} \text { Let } M&=(P-1 A P-I)^{2} \\ &=\left(P^{-1} A P\right)^{2}-2 P^{-1} A P+I \\ &=P^{-1} A^{2} P-2 P^{-1} A P+I \end{array}$

$\begin{aligned} PM &= A ^{2} P -2 AP + P \\ &=\left( A ^{2}-2 A . I + I ^{2}\right) P \\ \Rightarrow & \operatorname{Det}( PM )=\operatorname{Det}\left(( A - I )^{2} \times P \right) \\ \Rightarrow & \text { DetP.DetM }=\operatorname{Det}( A - I )^{2} \times \operatorname{Det}( P ) \\ \Rightarrow & \text { Det } M =(\operatorname{Det}( A - I ))^{2} \end{aligned}$

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

$\begin{array}{l} \operatorname{Det}(A-I)=\left(w^{2}+w+w\right)+7(-w)+w^{3}=-6 w \\ \operatorname{Det}((A-I))^{2}=36 w^{2} \\ \Rightarrow \quad \alpha=36 \end{array}$

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Let and where and be the identity matrix of order 3. If the determinant of the matrix is then the value of is equal to ____________. Option: 1 -6 Option: 2 36 Option: 3 -36 Option: 4 6

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