---
title: "Linear system solvable for which λ"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:linear-system-solvable-for-which"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/linear-system-solvable-for-which
site: BCT Engineering
---

# Linear system solvable for which λ

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2068 Chaitra, 5 marks
- Video: [Numerical on System of Linear Equations || #numericals || #matrices || #engineeringMathematics](https://www.youtube.com/watch?v=M-357rYNSno) — ENGINEER'S ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/linear-system-solvable-for-which
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

For what value of λ do the equations x + y + z = 1, x + 4y + 10z = λ² and x + 2y + 4z = λ have a solution? Solve them completely in each case.

## Short answer

Solutions exist only for λ = 1 or λ = 2. For λ=1: x = 1+2k, y = −3k, z = k; for λ=2: x = 2k, y = 1−3k, z = k (k arbitrary).

## Answer

The system is x + y + z = 1, x + 4y + 10z = λ², x + 2y + 4z = λ. It has a solution only if rank A = rank [A|B].

```text
[A|B] = [1 1  1 |  1 ]
        [1 4 10 | λ² ]
        [1 2  4 |  λ ]
```

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

```text
[1 1 1 |   1  ]
[0 3 9 | λ²−1 ]
[0 1 3 | λ−1  ]
```

**Step 2.** R₂ → R₂ − 3R₃:

```text
[1 1 1 |          1          ]
[0 0 0 | λ²−1−3λ+3 = λ²−3λ+2 ]
[0 1 3 |         λ−1         ]
```

Interchange R₂ and R₃:

```text
[1 1 1 |      1       ]
[0 1 3 |     λ−1      ]
[0 0 0 | (λ−1)(λ−2)   ]
```

#### Condition
rank A = 2 always. The system is consistent only if rank [A|B] = 2, i.e. (λ − 1)(λ − 2) = 0.

**The equations have a solution only for λ = 1 or λ = 2.** In both cases rank = 2 < 3, so there are infinitely many solutions (one free variable z = k).

#### Case λ = 1
- y + 3z = 0 ⇒ y = −3k
- x = 1 − y − z = 1 + 3k − k = 1 + 2k

**x = 1 + 2k, y = −3k, z = k.**

#### Case λ = 2
- y + 3z = 1 ⇒ y = 1 − 3k
- x = 1 − y − z = 1 − 1 + 3k − k = 2k

**x = 2k, y = 1 − 3k, z = k.**

For any other value of λ, the system is inconsistent (no solution).

## Check yourself

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

1. When does the system have a solution?
   A. When |A| ≠ 0
   B. When rank A = rank [A|B]
   C. When rank [A|B] = 3
   D. When rank A < rank [A|B]
   Answer: B. A system is consistent exactly when rank A equals rank of the augmented matrix. (Key point: Consistency condition)

2. After reduction, the zero row of A has right-hand side (λ−1)(λ−2). Which λ give a solution?
   A. λ = 1 or 2
   B. λ = 0 or 3
   C. λ = −1 or −2
   D. only λ = 1
   Answer: A. Rank A = 2, so the augmented rank must also be 2, requiring (λ−1)(λ−2)=0. (Key point: Row-reduced last row)

3. For a valid λ, how many solutions are there?
   A. Exactly one
   B. None
   C. Infinitely many, with two free variables
   D. Infinitely many, with one free variable
   Answer: D. Rank 2 < 3 unknowns gives 3−2 = 1 free variable. (Key point: Number of solutions)

4. With z = k and λ = 2, what is the solution?
   A. x = 1+2k, y = −3k, z = k
   B. x = 2k, y = 3k−1, z = k
   C. x = 2k, y = 1−3k, z = k
   D. x = 1−2k, y = 3k, z = k
   Answer: C. y + 3z = 1 gives y = 1−3k, and x = 1 − y − z = 2k. (Key point: Solution for λ = 2)

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