Matrices & Determinants
System of Linear Equations
Grade Class 12
Question:
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 homogeneous equations to have non-trivial solutions, the determinant of the coefficient matrix must be zero. Rearrange the equations to form a system: x - ay - az = 0, -bx + y - bz = 0, -cx - cy + z = 0. The determinant of the coefficient matrix is 1 - ab - bc - ca - 2abc = 0. Use this to evaluate the expression.
The given equations are: x - ay - az = 0, -bx + y - bz = 0, -cx - cy + z = 0. For non-trivial solutions, the determinant must be zero: |1 -a -a; -b 1 -b; -c -c 1| = 0. Expanding this gives 1(1-bc) + a(-b-bc) - a(bc+c) = 0, which simplifies to 1 - bc - ab - abc - abc - ac = 0, or 1 - (ab + bc + ca) - 2abc = 0. The expression to evaluate is 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 1 = ab+bc+ca+2abc, the expression evaluates to 2.
Correct Answer: 2