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 + i√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 ______________.

Solution

Maths sir · 4 steps

Pull P out of the bracket first. Then the big matrix P drops away and you only need det(A - I).

1. Factor out P. Since I = P⁻¹IP, you can write P⁻¹AP - I = P⁻¹(A - I)P. The determinant of a product is the product of the determinants, and det P⁻¹ × det P = 1.

   det[(P⁻¹AP - I)²] = [det(P⁻¹(A-I)P)]² = [det(A - I)]²

2. Build A minus I. Subtract 1 from each diagonal entry. Use 1 + ω + ω² = 0, so -ω - 1 = ω², and -ω + 1 - 1 = -ω. This gives A - I = (1, 7, ω²; -1, ω², 1; 0, -ω, -ω).
3. Expand column 1. The first entry gives 1 × (-ω³ + ω). The second gives (-1)(-1)(-7ω + ω³) = -7ω + ω³. The ω³ terms cancel, so det(A - I) = ω - 7ω = -6ω.
4. So. The determinant is (-6ω)² = 36ω². Matching this with αω² gives α = 36.

The idea to keep. Whenever you see P⁻¹AP, remember that P⁻¹AP and A have the same determinant. Write I as P⁻¹IP and the ugly matrix P never needs to be touched.

Answer from the NTA final answer key. Solution by PrepWiser.