Question

Let P = ${\displaystyle \left[\begin{array}{ccc}-30& 20& 56\\ 90& 140& 112\\ 120& 60& 14\end{array}\right]}$ and A = ${\displaystyle \left[\begin{array}{ccc}2& 7& {\omega }^2\\ -1& -\omega & 1\\ 0& -\omega & -\omega +1\end{array}\right]}$ where ω = ${\displaystyle \frac{-1+i\sqrt{3}}{2}}$, and I₃ be the identity matrix of order 3. If the determinant of the matrix (P^–1AP – I₃)² is αω², then the value of α is equal to ______.

Options

• 35

• 36

• 37

• 38

Answer 1

Let P = ${\displaystyle \left[\begin{array}{ccc}-30& 20& 56\\ 90& 140& 112\\ 120& 60& 14\end{array}\right]}$ and A = ${\displaystyle \left[\begin{array}{ccc}2& 7& {\omega }^2\\ -1& -\omega & 1\\ 0& -\omega & -\omega +1\end{array}\right]}$ where ω = ${\displaystyle \frac{-1+i\sqrt{3}}{2}}$, and I₃ be the identity matrix of order 3. If the determinant of the matrix (P^–1AP – I₃)² is αω², then the value of α is equal to 36.

Explanation:

Given that P = ${\displaystyle \left[\begin{array}{ccc}-30& 20& 56\\ 90& 140& 112\\ 120& 60& 14\end{array}\right]}$

A = ${\displaystyle \left[\begin{array}{ccc}2& 7& {\omega }^2\\ -1& -\omega & 1\\ 0& -\omega & -\omega +1\end{array}\right]}$; ω = ${\displaystyle \frac{-1+i\sqrt{3}}{2}}$

Let M =  (P^–1AP – I₃)² 

⇒ M = (P^–1AP)² + (I₃)² – 2(P^–1AP)(I₃)

M = (P^–1AP)²P + I₃ – 2(P^–1AP)

M = P^–1A²P + I₃ – 2P^–1AP

PM = A²P + PI₃ – 2AP

⇒ PM = (A² + I₃ – 2A)P

PM = (A² + (I₃)² – 2AI₃)P

⇒ PM = (A – I₃)²P

 Det(PM) = Det((A – I₃)²P)

(Det P)(Det M) = (Det(A – I₃)²)(Det P)

Det M = Det(A – I₃)²

Now, A – I₃ = ${\displaystyle \left[\begin{array}{ccc}2& 7& {\omega }^2\\ -1& -\omega & 1\\ 0& -\omega & -\omega +1\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$

A – I₃ = ${\displaystyle \left[\begin{array}{ccc}1& 7& {\omega }^2\\ -1& -\omega -1& 1\\ 0& -\omega & -\omega \end{array}\right]}$

Det(A – I₃) = 1(ω² + ω + ω) – 7(ω – 0) + ω²(ω – 0)

⇒ Det(A – I₃) = ω² + 2ω – 7ω + ω³

⇒ Det(A – I₃) = ω³ + ω² – 5ω

⇒ Det(A – I₃) = –6ω

⇒ Det(A – I₃)² = 36ω²

⇒ αω² = 36ω²

⇒ α = 36.