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What values of x, y and z satisfy the following system of linear equations? [1 2 3] [x] [6 ] [2 1 2] [y] = [10] [1 1 1] [z] [8 ] A. x = 6, y = 3, z = 2 B. x = 12, y = 3, z = -4 C. x = 6, y = 6, z = -4 D. x = 12, y = -3, z = 0

GATE 2004 · Engineering Mathematics · System of Equations · easy

Answer: C. x = 6, y = 6, z = -4.

  1. Form augmented matrix and eliminate x: Augmented matrix: [[1,2,3|6],[2,1,2|10],[1,1,1|8]]. Apply R2 = R2 - 2*R1: [2-2, 1-4, 2-6 | 10-12] = [0, -3, -4 | -2]. Apply R3 = R3 - R1: [1-1, 1-2, 1-3 | 8-6] = [0, -1, -2 | 2]. Result: [[1,2,3|6],[0,-3,-4|-2],[0,-1,-2|2]].
  2. Eliminate y from equation 2: Apply R2 = R2 - 3*R3: [0, -3-(-3), -4-(-6) | -2-6] = [0, 0, 2 | -8]. The system is now upper triangular: [[1,2,3|6],[0,-1,-2|2],[0,0,2|-8]].
  3. Back substitution to find z, y, x: From row 3: 2z = -8, so z = -4. From row 2 (R3 after swap): -y - 2(-4) = 2 => -y + 8 = 2 => y = 6. From row 1: x + 2(6) + 3(-4) = 6 => x + 12 - 12 = 6 => x = 6. Solution: x = 6, y = 6, z = -4.