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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2021 · 16 March · Morning Shift · Mathematics · Matrices

Let P = [ cases - 30 & 20 & 56 90 & 140 & 112 120 & 60 & 14 cases ]...

Let P = \left[\begin{cases}- 30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14\end{cases}\right] andA = \left[{27ω2−1−ω10−ω−ω+1\begin{cases}2 & 7 & {\omega}^{2} \\ - 1 & - \omega & 1 \\ 0 & - \omega & - \omega + 1\end{cases}\right]wherewhere\omega = \dfrac{- 1 + i\sqrt{3}}{2},and, and{I}_{3}betheidentitymatrixoforderbe the identity matrix of order3.Ifthedeterminantofthematrix. If the determinant of the matrix{\left({P}^{- 1}AP - {I}_{3}\right)}^{2} is αω2\alpha{\omega}^{2}, then the value of α\alpha is equal to _________.

Correct answer

0

Step-by-step solution

Let M=(P−1AP−I)2M = {\left({P}^{- 1}AP - I\right)}^{2}

=(P−1AP)2−2P−1AP+I= {\left({P}^{- 1}AP\right)}^{2} - 2{P}^{- 1}AP + I

=P−1A2P−2P−1AP+I(∵(P−1AP)2=(P−1AP)(P−1AP))= {P}^{- 1}{A}^{2}P - 2{P}^{- 1}AP + I\left(∵ {\left({P}^{- 1}AP\right)}^{2} = \left({P}^{- 1}AP\right)\left({P}^{- 1}AP\right)\right)

Multiplying with PP on both sides

⇒PM=A2P−2AP+P⇒ PM = {A}^{2}P - 2AP + P

=(A2−2A.I+I2)P= \left({A}^{2} - 2A . I + {I}^{2}\right)P

⇒Det(PM)=Det((A−I)2×P)⇒ Det \left(PM\right) = Det \left({\left(A - I\right)}^{2} \times P\right)

⇒Det(PM)=Det(A−I)2×Det(P)⇒ Det \left(PM\right) = Det {\left(A - I\right)}^{2} \times Det \left(P\right)

⇒DetM=(Det(A−I))2⇒ DetM = {\left(Det\left(A - I\right)\right)}^{2}

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

Det \left(A - I\right) = \left({w}^{2} + w + w\right) + 7\left(- w\right) + {w}^{3} = - 6w</p><p>Det((A−I))2=36w2</p><p>Det {\left(\left(A - I\right)\right)}^{2} = 36{w}^{2}

⇒α=36⇒ \alpha = 36

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