If the equations a(y + z) = x, b(z + x) = y, c(x + y) = z (where a, b, c ≠ -1) have nontrivial solutions, then find the value of 1/(1+a) + 1/(1+b) + 1/(1+c).
Step-by-Step Solution
Key Concept: For a system of linear equations to have nontrivial solutions, the determinant of the coefficient matrix must be zero. Rewrite the equations as -x + ay + az = 0, bx - y + bz = 0, cx + cy - z = 0. The determinant of the matrix [[-1, a, a], [b, -1, b], [c, c, -1]] = 0.
The given system of equations is: -x + ay + az = 0, bx - y + bz = 0, cx + cy - z = 0. For nontrivial solutions, the determinant of the coefficient matrix must be zero: det([[-1, a, a], [b, -1, b], [c, c, -1]]) = 0. Expanding the determinant: -1(1 - bc) - a(-b - bc) + a(bc + c) = 0 => -1 + bc + ab + abc + abc + ac = 0 => ab + bc + ca + 2abc = 1. Dividing by (1+a)(1+b)(1+c) = 1 + a + b + c + ab + bc + ca + abc, we can relate this to the expression 1/(1+a) + 1/(1+b) + 1/(1+c) = ((1+b)(1+c) + (1+a)(1+c) + (1+a)(1+b)) / ((1+a)(1+b)(1+c)) = (3 + 2(a+b+c) + ab+bc+ca) / (1 + a+b+c + ab+bc+ca + abc). Using the condition, the value evaluates to 2.
Correct Answer: 2