A determinant g(y) is defined as: g(y) = |p-y q r | |r p-y q | |q r p-y| where p q r 0. If y1, y2, y3 are the cube roots of unity (roots of y³ = 1) and f(y) = p + q*y + r*y², find g(0).

Question

Accepted Answer

f(y1) * f(y2) * f(y3). The matrix with rows (p-y, q, r), (r, p-y, q), (q, r, p-y) is a circulant matrix. Its determinant equals the product (a + b*w + c*w²)(a + b*w² + c*w⁴)(a + b + c) where w = e^(2*pi*i/3) is a primitive cube root of unity, and here a = p-y, b = q, c = r. At y=0: g(0) = [(p + q*w + r*w²)(p + q*w² + r*w⁴)(p + q + r)] where w, w², 1 are the cube roots of unity. So g(0) = f(w)*f(w²)*f(1) = f(y1)*f(y2)*f(y3) where f(y) = p + q*y + r*y².

A determinant g(y) is defined as: g(y) = |p-y q r | |r p-y q | |q r p-y| where p > q > r > 0. If y1, y2, y3 are the cube roots of unity (roots of y³ = 1) and f(y) = p + qy + ry², find g(0).

Solution

Related JEE Advanced Maths questions

A determinant g(y) is defined as: g(y) = |p-y q r | |r p-y q | |q r p-y| where p > q > r > 0. If y1, y2, y3 are the cube roots of unity (roots of y³ = 1) and f(y) = p + q*y + r*y², find g(0).

Solution

Related JEE Advanced Maths questions

A determinant g(y) is defined as: g(y) = |p-y q r | |r p-y q | |q r p-y| where p > q > r > 0. If y1, y2, y3 are the cube roots of unity (roots of y³ = 1) and f(y) = p + qy + ry², find g(0).