--- title: "Consistency of a linear system" subject: "Mathematics III" chapter: 1 key: "semester_3/mathematics_iii:consistency-of-a-linear-system-seven" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/consistency-of-a-linear-system-seven site: BCT Engineering --- # Consistency of a linear system Mathematics III · Chapter 1: Determinants and Matrices - Asked in: 2071 Chaitra, 5 marks - Video: [Consistency of system of linear equations using rank method](https://www.youtube.com/watch?v=povS4KL3nR8) — maths voice - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/consistency-of-a-linear-system-seven - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1 ## Question Test the consistency of the system x − 6y − z = 10, 2x − 2y + 3z = 10, 3x − 8y + 2z = 20 and solve completely, if found consistent. ## Short answer Reduce augmented matrix to echelon form; rank A = rank [A|B] = 2 < 3, consistent with infinitely many solutions. Solution: x = 4 − 2k, y = −1 − k/2, z = k. ## Answer A system AX = B is consistent if rank A = rank [A|B]. ```text [A|B] = [1 −6 −1 | 10] [2 −2 3 | 10] [3 −8 2 | 20] ``` **Step 1.** R₂ → R₂ − 2R₁, R₃ → R₃ − 3R₁: ```text [1 −6 −1 | 10] [0 10 5 | −10] [0 10 5 | −10] ``` **Step 2.** R₃ → R₃ − R₂: ```text [1 −6 −1 | 10] [0 10 5 | −10] [0 0 0 | 0] ``` #### Consistency rank A = 2 = rank [A|B]. The system is **consistent**. As the rank (2) is less than the number of unknowns (3), it has **infinitely many solutions** depending on one parameter. #### Solution Let z = k. - 10y + 5k = −10 ⇒ y = −1 − k/2 - x − 6y − z = 10 ⇒ x = 10 + 6(−1 − k/2) + k = 4 − 2k **Answer: x = 4 − 2k, y = −1 − k/2, z = k, for any real k.** Check with k = 2: (0, −2, 2): 0 + 12 − 2 = 10 ✓; 0 + 4 + 6 = 10 ✓; 0 + 16 + 4 = 20 ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After R2 - 2R1 and R3 - 3R1, rows 2 and 3 are: A. Different B. Both zero C. Identical (0, 10, 5 | -10) D. Negatives of each other Answer: C. Both give (0, 10, 5 | -10), so one becomes zero. (Key point: Step 1) 2. The ranks are: A. rank A = 2 = rank [A|B] B. rank A = 2, rank [A|B] = 3 C. rank A = 3 = rank [A|B] D. rank A = 1 Answer: A. Equal ranks mean consistent. (Key point: Consistency) 3. Since rank 2 < 3 unknowns: A. Unique solution B. No solution C. Infinitely many solutions depending on one parameter D. Only the zero solution Answer: C. One free variable. (Key point: Solution type) 4. With z = k, y equals: A. -1 + k/2 B. 1 + k/2 C. -1 - 2k D. -1 - k/2 Answer: D. 10y + 5k = -10, so y = -1 - k/2, and x = 4 - 2k. 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