Consider the following set of linear equations
x₁ + x₂ + x₃ = 6
2x₁ + 2x₂ + 3x₃ = 14
3x₁ + x₂ + 2x₃ = 14
The solution for this set exists only when the value of x₂ is _________.
Official Answer Key:
1 to 1
Step-by-Step Solution & Explanation
Coefficient-matrix determinant:
|1 1 1; 2 2 3; 3 1 2| = 1(2·2−3·1) − 1(2·2∓3·3) + 1(2·1−2·3) = 1(1) − 1(−5) + 1(−4) = 1+5−4 = 2 ≠ 0,
so a unique solution exists. Solve directly: Eq2 − 2×Eq1 gives x₃ = 2. Substituting into Eq1: x₁+x₂ = 4. Substituting x₃=2 into Eq3: 3x₁+x₂ = 10. Solving these two: x₁ = 3, x₂ = 1, x₃ = 2 (consistent with all three original equations).
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