1. Tardigrade
  2. Question
  3. Mathematics
  4. Let P=[-30 20 56 90 140 112 120 60 14] and A=[2 7 ω2 -1 -ω 1 0 -ω -ω+1] where ω=(-1+ i √3/2), and I 3 be the identity matrix of order 3 . If the determinant of the matrix ( P -1 AP - I 3)2 is α ω2, then the value of α is equal to .

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Q. 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 ______.

JEE MainJEE Main 2021Matrices

Solution:

Let $M =\left( P ^{-1} AP - I \right)^{2}$
$=\left( P ^{-1} AP \right)^{2}-2 P ^{-1} AP + I$
$= P ^{-1} A ^{2} P -2 P ^{-1} AP + I$
$PM = A ^{2} P -2 AP + P$
$=\left( A ^{2}-2 A \cdot I + I ^{2}\right) P$
$\Rightarrow \operatorname{Det}( PM )=\operatorname{Det}\left(( A - I )^{2} \times P \right)$
$\Rightarrow $ DetP.DetM $=\operatorname{Det}( A - I )^{2} \times \operatorname{Det}( P )$
$\Rightarrow $ Det $M =(\operatorname{Det}( A - I ))^{2}$
Now $A - I =\begin{bmatrix}1 & 7 & w ^{2} \\ -1 & - w -1 & 1 \\ 0 & - w & - w \end{bmatrix}$
$\operatorname{Det}( A - I )=\left( w ^{2}+ w + w \right)+7(- w )+ w ^{3}=-6 u$
$\operatorname{Det}(( A - I ))^{2}=36 w ^{2}$
$\Rightarrow \alpha=36$