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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2079 Bhadra, 5 marks
- Video: [Linear Algebra : Proofs for Determinant Identities](https://www.youtube.com/watch?v=jcb3o4MITgg) — Maths Resource
- 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-proof-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Prove that [a³ 3a² 3a 1; a² a²+2a 2a+1 1; a 2a+1 a+2 1; 1 3 3 1] (determinant) = (a − 1)⁶ by using properties of determinants.

## Short answer

Perform R1−R2, R2−R3, R3−R4, factor (a−1)³, and expand along the fourth column. The resulting 3×3 determinant reduces to (a−1)³, giving the total (a−1)⁶.

## Answer

Let

```text
Δ = | a³   3a²     3a    1 |
    | a²  a²+2a   2a+1   1 |
    | a   2a+1    a+2    1 |
    | 1    3       3     1 |
```

**Step 1: R₁ → R₁ − R₂, R₂ → R₂ − R₃, R₃ → R₃ − R₄.**

- R₁: (a³ − a², 3a² − a² − 2a, 3a − 2a − 1, 0) = (a²(a−1), 2a(a−1), (a−1), 0)
- R₂: (a² − a, a² + 2a − 2a − 1, 2a + 1 − a − 2, 0) = (a(a−1), (a−1)(a+1), (a−1), 0)
- R₃: (a − 1, 2a + 1 − 3, a + 2 − 3, 0) = ((a−1), 2(a−1), (a−1), 0)

**Step 2: take (a − 1) common from each of R₁, R₂, R₃** (factor (a − 1)³):

```text
Δ = (a−1)³ | a²   2a    1   0 |
           | a   a+1    1   0 |
           | 1    2     1   0 |
           | 1    3     3   1 |
```

**Step 3: expand along C₄**, which has the single non-zero entry 1 at position (4, 4) (sign +):

```text
Δ = (a−1)³ | a²  2a   1 |
           | a   a+1  1 |
           | 1   2    1 |
```

**Step 4: R₁ → R₁ − R₃, R₂ → R₂ − R₃.**

```text
| a²−1   2a−2   0 |
| a−1    a−1    0 |
|  1      2     1 |
```

Expanding along C₃ (entry 1 at (3, 3)):

```text
(a²−1)(a−1) − (2a−2)(a−1) = (a−1)²[(a+1) − 2] = (a−1)³
```

Therefore Δ = (a−1)³ · (a−1)³ = **(a − 1)⁶.** ∎

## Check yourself

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

1. The first step is
   A. R1->R1-R2, R2->R2-R3, R3->R3-R4
   B. C1 -> C1 - C2
   C. R4->R4-R1
   D. expand along R1
   Answer: A. This makes the last column (0,0,0,1). (Key point: Row operation)

2. Which factor is taken out?
   A. (a-1) once
   B. (a-1) from each of three rows, giving (a-1)^3
   C. (a+1)^3
   D. a^3
   Answer: B. Each of R1, R2, R3 contains a-1. (Key point: Common factor)

3. After two more reductions the 3x3 determinant equals
   A. (a-1)^2
   B. (a+1)^3
   C. (a-1)^3, so Delta = (a-1)^6
   D. a^3-1
   Answer: C. (a-1)^3 times (a-1)^3. (Key point: Final)

## Close variants set in other exams

- [Determinant identity proof](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-proof-five) — Mathematics III, chapter 1, asked 9×
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