Let $\mathrm{P}=\left[\begin{array}{ccc}-30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ccc}2 & 7 & \omega^{2} \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1\end{array}\right]$ where $\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2}$, and $\mathrm{I}_{3}$ be the identity matrix of order 3. If the determinant of the matrix $\left(\mathrm{P}^{-1} \mathrm{AP}-\mathrm{I}_{3}\right)^{2}$ is $\alpha \omega^{2}$, then the value of $\alpha$ is equal to

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Matrices

Let

P = - 30 90120 20140605611214

and

A = 2 - 1 0 7 - ω - ω ω^{2} 1 - ω + 1

where

ω = \frac{- 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}

α ω^{2}

, then the value of

α

is equal to

Answer: 36

Similar matrices have equal determinants, so

det ((P^{- 1} A P - I_{3})^{2}) = (det (P^{- 1} (A - I_{3}) P))^{2} = (det (A - I_{3}))^{2}

Since

1 + ω + ω^{2} = 0

gives

- ω - 1 = ω^{2}

A - I_{3} = 1 - 1 0 7 ω^{2} - ω ω^{2} 1 - ω

Expanding along the first column, the two surviving cofactors give

det (A - I_{3}) = ω^{2} - ω 1 - ω + 7 - ω ω^{2} - ω = (ω - ω^{3}) + (ω^{3} - 7 ω) = - 6 ω

Hence

(det (A - I_{3}))^{2} = 36 ω^{2}

, so

α = 36

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer