---
title: "Consistency of a linear system"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:consistency-of-a-linear-system-nine"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/consistency-of-a-linear-system-nine
site: BCT Engineering
---

# Consistency of a linear system

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2079 Baishakh, 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-nine
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the rank of the augmented matrix and test the consistency of the system of linear equations x + 9y − z = 27, x − 8y + 16z = 10, 2x + y + 15z = 37. Also find the solution if the system is consistent.

## Short answer

The augmented matrix has rank 2, equal to rank A, so the system is consistent with infinitely many solutions. Let z = k; then y = 1 + k, x = 18 − 8k.

## Answer

The system is x + 9y − z = 27, x − 8y + 16z = 10, 2x + y + 15z = 37. It is consistent if rank A = rank [A|B].

```text
[A|B] = [1  9 −1 | 27]
        [1 −8 16 | 10]
        [2  1 15 | 37]
```

**Step 1.** R₂ → R₂ − R₁, R₃ → R₃ − 2R₁:

```text
[1   9  −1 |  27]
[0 −17  17 | −17]
[0 −17  17 | −17]
```

**Step 2.** R₃ → R₃ − R₂, and divide R₂ by −17:

```text
[1  9 −1 | 27]
[0  1 −1 |  1]
[0  0  0 |  0]
```

#### Rank and consistency
- Rank of the augmented matrix [A|B] = 2
- Rank of A = 2

The ranks are equal, so the system is **consistent**. Since rank 2 < 3 unknowns, there are **infinitely many solutions** (one free variable).

#### Solution
Let z = k.
- y − z = 1 ⇒ y = 1 + k
- x + 9y − z = 27 ⇒ x = 27 − 9(1 + k) + k = 18 − 8k

**Answer: x = 18 − 8k, y = 1 + k, z = k, for any real k.**

Check with k = 1: (10, 2, 1): 10 + 18 − 1 = 27 ✓; 10 − 16 + 16 = 10 ✓; 20 + 2 + 15 = 37 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. After R₂−R₁ and R₃−2R₁, rows 2 and 3 are:
   A. Zero
   B. Identical (0 −17 17 | −17)
   C. Different
   D. (0 17 −17 | 17) and (0 0 0 | 0)
   Answer: B. Both become (0,−17,17|−17). (Key point: Elimination)

2. Rank A and rank [A|B] are:
   A. Both 2
   B. Both 3
   C. 2 and 3
   D. 1 and 2
   Answer: A. Two non-zero rows remain in both. (Key point: Ranks)

3. Since rank 2 < 3 unknowns:
   A. Unique solution
   B. No solution
   C. Two solutions
   D. Infinitely many solutions
   Answer: D. One free variable. (Key point: Type of solution)

4. Solution with z=k:
   A. x = 10, y = 2, z = 1 only
   B. x = 18+8k, y = 1−k, z = k
   C. x = 18−8k, y = 1+k, z = k
   D. x = 27−k, y = k, z = 1
   Answer: C. y = 1+k; x = 27−9(1+k)+k = 18−8k. (Key point: Solution)

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