The following system of equations: x_1 + 2x_2 = 1 3x_2 = 2 x_1 + 4x_2 + omega*x_3 = 4 has a unique solution. The only possible value(s) for omega is/are: A. 0 B. either 0 or 1 C. one of 0, 1, or -1 D. any real number

GATE 2008 · Engineering Mathematics · System of Equations · easy

Answer: The system has a unique solution for any omega != 0. Answer: D (any real number, interpreted as any nonzero real number). This question is marked excluded (X) in the official answer key.

1. Solve for x_2 from equation 2: From equation 2: x_2 = 2/3
2. Solve for x_1 from equation 1: x_1 = 1 - 2*(2/3) = 1 - 4/3 = -1/3
3. Analyze third equation for x_3: omega * x_3 = 4 - (-1/3) - 4*(2/3) = 4 + 1/3 - 8/3 = 4 - 7/3 = 5/3. So x_3 = 5/(3*omega) if omega != 0.
4. Conclusion on omega: The system has a unique solution for any omega != 0. The question answer is D (any real number) as noted; however technically omega must be nonzero. The answer key marks this as X (excluded/disputed). The correct answer intended is D.