Question

Let P = `[(-30, 20, 56),(90, 140, 112),(120, 60, 14)]` and A = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)]` where ω = `(-1 + isqrt(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 ______.

पर्याय

• 35

• 36

• 37

• 38

Answer 1

Let P = `[(-30, 20, 56),(90, 140, 112),(120, 60, 14)]` and A = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)]` where ω = `(-1 + isqrt(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 = `[(-30, 20, 56),(90, 140, 112),(120, 60, 14)]`

A = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)]`; ω = `(-1 + isqrt(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 = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)] - [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

A – I₃ = `[(1, 7, ω^2),(-1, -ω -1, 1),(0, -ω, -ω)]`

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.