Question

Let $P=\left[\begin{array}{ccc}-30& 20& 56\\ 90& 140& 112\\ 120& 16& 14\end{array}\right]$ and $A=\left[\begin{array}{ccc}2& 7& {\omega }^2\\ -1& -\omega & 1\\ 0& -\omega & -\omega +1\end{array}\right]$

where $𝛚=\frac{(-1+i\sqrt{3})}{2}$, and $I_3$ be the identity matrix of order $3.$

If the determinant of the matrix $\begin{array}{rcl}& & {\left|P^{1}AP-1\right|}^2\end{array}$ is $\begin{array}{rcl}& & \alpha {\omega }^2\end{array}$, then the value of $\alpha$ is equal to

Answer 1

The value of $\alpha =36$

Finding the value of $\alpha$

Simplifying the required expression value using given parameters, we have

$\begin{array}{rcl}{\left|P^{1}AP-1\right|}^2& =& \alpha {\omega }^2\\ & =& \left|(P^{-1}AP-1)(P^{-1}AP-1)\right|\\ & =& \left|P^{-1}AP{P}^{-1}AP-2P^{-1}AP+I\right|\\ & =& \left|P^{-1}{A}^{2}P-2P^{-1}AP+P^{-1}P\right|\\ & =& \left|P^{-1}(A^2-2A+I)P\right|\\ & =& \left|P^{-1}{(A-I)}^{2}P\right|\\ & =& \left|P^{-1}\right|{\left|(A-I)\right|}^2\left|P\right|\\ & =& {\left|(A-I)\right|}^2\\ & =& {\left|\left[\begin{array}{ccc}2& 7& {\omega }^2\\ -1& -\omega & 1\\ 0& -\omega & -\omega +1\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\right|}^2\\ & =& {\left[\begin{array}{ccc}1& 7& {\omega }^2\\ -1& -\omega -1& 1\\ 0& -\omega & -\omega \end{array}\right]}^2\\ & =& \left(1\right({\omega }^2+\omega +\omega )-7(-\omega )+{\omega }^2(\omega )\\ & =& ({\omega }^2+2\omega -7\omega +1{)}^2\left[{\omega }^3=1\right]\\ & =& {({\omega }^2–5\omega +1)}^2\left[(1+\omega +{\omega }^2)=0\Rightarrow (1+{\omega }^2)=-{\omega }^2\right]\\ \therefore \alpha {\omega }^2& =& {(-6\omega )}^2=36{\omega }^2\\ & \Rightarrow & ɑ=36\end{array}$

Hence, The value of $\alpha =36$