Let $$P = \left[ {\matrix{
{ - 30} & {20} & {56} \cr
{90} & {140} & {112} \cr
{120} & {60} & {14} \cr
} } \right]$$ and
$$A = \left[ {\matrix{
2 & 7 & {{\omega ^2}} \cr
{ - 1} & { - \omega } & 1 \cr
0 & { - \omega } & { - \omega + 1} \cr
} } \right]$$ where
$\omega = {{ - 1 + i\sqrt 3 } \over 2}$, and I3 be the identity matrix of order 3. If the
determinant of the matrix (P$-$1AP$-$I3)2 is $\alpha$$\omega$2, then the value of $\alpha$ is equal to ______________.
Answer (integer)
36
Solution
$|{P^{ - 1}}AP - I{|^2}$<br><br>$= |({P^{ - 1}}AP - I){({P^{ - 1}}AP - 1)^2}|$<br><br>$= |{P^{ - 1}}AP{P^{ - 1}}AP - 2{P^{ - 1}}AP + I|$<br><br>$= |{P^{ - 1}}{A^2}P - 2{P^{ - 1}}AP + {P^{ - 1}}IP|$<br><br>$= |{P^{ - 1}}({A^2} - 2A + I)P|$<br><br>$= |{P^{ - 1}}{(A - I)^2}P|$<br><br>$= |{P^{ - 1}}||A - I{|^2}|P|$<br><br>$= |A - I{|^2}$<br><br>$$ = \left| {\matrix{
1 & 7 & {{\omega ^2}} \cr
{ - 1} & { - \omega - 1} & 1 \cr
0 & { - \omega } & { - \omega } \cr
} } \right|$$<br><br>$= {(1(\omega (\omega + 1) + \omega ) - 7\omega + {\omega ^2}.\omega )^2}$<br><br>$= {({\omega ^2} + 2\omega - 7\omega + 1)^2}$<br><br>$= {({\omega ^2} - 5\omega + 1)^2}$<br><br>$= {( - 6\omega )^2}$<br><br>$= 36{\omega ^2}$
<br><br>$\therefore$ $\alpha$$\omega$<sup>2</sup> = $36{\omega ^2}$
<br><br>$\Rightarrow \alpha = 36$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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