Let a, b, c, l, m, n ∈ R such that al + bm + cn = 0, bl + cm + an = 0, cl + am + bn = 0. If a, b & c are distinct & f(x) = ax^3 + bx^2 + cx + 5, then the value of f(1) is
Step-by-Step Solution
Key Concept: The given system of equations can be written as a matrix equation AX = 0 where X = [l, m, n]^T. Since l, m, n are not all zero (implied by the structure), the determinant of the coefficient matrix must be zero. This leads to a condition on a, b, c. Since a, b, c are distinct, this forces a = b = c = 0, which contradicts the distinct condition, or implies a specific relation. Actually, the determinant of the matrix [a b c; b c a; c a b] = 0 implies -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0. Since a, b, c are distinct, a^2+b^2+c^2-ab-bc-ca is not zero, so a+b+c = 0. Then f(1) = a+b+c+5 = 0+5 = 5.
The system of equations is: <br> al + bm + cn = 0 <br> bl + cm + an = 0 <br> cl + am + bn = 0 <br> This is a homogeneous system of linear equations in l, m, n. For non-trivial solutions, the determinant of the coefficient matrix must be zero: <br> |a b c; b c a; c a b| = 0 <br> -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0 <br> Since a, b, c are distinct, a^2+b^2+c^2-ab-bc-ca = 1/2((a-b)^2+(b-c)^2+(c-a)^2) \neq 0. <br> Therefore, a+b+c = 0. <br> Now, f(1) = a(1)^3 + b(1)^2 + c(1) + 5 = a+b+c+5. <br> Substituting a+b+c = 0, we get f(1) = 0 + 5 = 5.
Correct Answer: 5