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 a, b, c are distinct, the determinant of the coefficient matrix is non-zero, implying the only solution is l=m=n=0. However, for non-trivial solutions, the determinant must be zero. Since the problem implies a, b, c are distinct, the only way for the system to hold is if a=b=c=0, but they are distinct. Re-evaluating, the determinant of the matrix [a b c; b c a; c a b] must be zero. This determinant is -(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: al + bm + cn = 0, bl + cm + an = 0, cl + am + bn = 0. This can be written as a matrix equation: [[a, b, c], [b, c, a], [c, a, b]] * [l, m, n]^T = [0, 0, 0]^T. For non-trivial solutions (l, m, n), the determinant of the coefficient matrix must be zero. The determinant is -(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 = 1/2 * ((a-b)^2 + (b-c)^2 + (c-a)^2) is not zero. Thus, a+b+c = 0. We need to find f(1) = a(1)^3 + b(1)^2 + c(1) + 5 = a+b+c+5. Substituting a+b+c=0, we get f(1) = 0+5 = 5.
Correct Answer: 5