---
title: "Determinant identity: 1"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:determinant-identity-1"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-1
site: BCT Engineering
---

# Determinant identity: 1

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2078 Kartik, 5 marks; 2075 Chaitra, 5 marks; 2074 Chaitra, 5 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

If [a a² a³−1; b b² b³−1; c c² c³−1] (determinant) = 0, where a ≠ b ≠ c, apply the properties of determinants to show abc = 1.

## Short answer

Split the determinant into two: one equals abc times the Vandermonde determinant V, the other equals V. Thus Δ = (abc − 1)V. Since a,b,c are distinct, V ≠ 0, so Δ = 0 implies abc = 1.

## Answer

Let Δ be the determinant:

```text
Δ = | a  a²  a³−1 |
    | b  b²  b³−1 |
    | c  c²  c³−1 |
```

#### Step 1: Split the third column
Each entry of C₃ is a difference (a³ − 1), so by the additive property of determinants:

```text
Δ = | a a² a³ |   −   | a a² 1 |
    | b b² b³ |       | b b² 1 |
    | c c² c³ |       | c c² 1 |
```

#### Step 2: Evaluate the first determinant
Take a, b, c common from R₁, R₂, R₃:

```text
| a a² a³ |           | 1 a a² |
| b b² b³ | = abc  ·  | 1 b b² | = abc · V
| c c² c³ |           | 1 c c² |
```

where V is the Vandermonde determinant: V = (b − a)(c − a)(c − b).

#### Step 3: Evaluate the second determinant
Move C₃ to the first position: C₃ ↔ C₂, then C₂ ↔ C₁ (two interchanges, so the sign is unchanged):

```text
| a a² 1 |   | 1 a a² |
| b b² 1 | = | 1 b b² | = V
| c c² 1 |   | 1 c c² |
```

#### Step 4: Combine
```text
Δ = abc · V − V = (abc − 1) V = (abc − 1)(b − a)(c − a)(c − b)
```

Given Δ = 0 and a ≠ b ≠ c (all distinct, so V ≠ 0), we must have

**abc − 1 = 0, i.e. abc = 1.** ∎

## Check yourself

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

1. Why can the determinant be split into two?
   A. Because the rows are equal
   B. Each entry of C3 is a difference (a^3 - 1)
   C. Because the determinant is zero
   D. Because C3 is the sum of C1 and C2
   Answer: B. Additive property: the column of differences gives a difference of determinants. (Key point: Splitting the column)

2. The first determinant equals abc times what?
   A. The Vandermonde V = (b-a)(c-a)(c-b)
   B. abc
   C. (a+b+c)
   D. a^2 b^2 c^2
   Answer: A. Taking a, b, c from the rows leaves the Vandermonde determinant. (Key point: Vandermonde)

3. What is Delta after combining?
   A. (abc + 1)V
   B. abc V
   C. (abc - 1)(b-a)(c-a)(c-b)
   D. (abc - 1) + V
   Answer: C. Delta = abc V - V. (Key point: Result)

4. Why does Delta = 0 give abc = 1?
   A. Because V = 0
   B. Because a = b = c
   C. Because abc = 0
   D. a, b, c distinct so V != 0
   Answer: D. Since V is non-zero, abc - 1 must be zero. (Key point: Conclusion)

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