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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2068 Chaitra, 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-twelve
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Prove that [a b c; b c a; c a b]² (determinant squared) = [2bc−a² c² b²; c² 2ac−b² a²; b² a² 2ab−c²] (determinant) = (a³ + b³ + c³ − 3abc)².

## Answer

Let D = |a b c; b c a; c a b| and let Δ be the determinant on the right-hand side:

```text
Δ = | 2bc−a²   c²      b²   |
    |  c²    2ca−b²    a²   |
    |  b²      a²    2ab−c² |
```

#### Step 1: Evaluate D
Add C₂ and C₃ to C₁: every entry of C₁ becomes s = a + b + c.

```text
D = s | 1  b  c |
      | 1  c  a |
      | 1  a  b |
```

R₂ → R₂ − R₁, R₃ → R₃ − R₁:

```text
D = s | 1  b    c   |
      | 0  c−b  a−c |
      | 0  a−b  b−c |
  = s[(c−b)(b−c) − (a−c)(a−b)]
  = −s[(b−c)² + (a−c)(a−b)]
  = −s(a² + b² + c² − ab − bc − ca)
  = −(a³ + b³ + c³ − 3abc)
```

So D² = (a³ + b³ + c³ − 3abc)², which proves the first and last members are equal.

#### Step 2: Show Δ = D²
Let G = [−a −b −c; c a b; b c a]. Multiply D by G (row by column):

```text
D·G = [a b c][−a −b −c]
      [b c a][ c  a  b]
      [c a b][ b  c  a]
```

- (1,1): −a² + bc + cb = 2bc − a²
- (1,2): −ab + ab + c² = c²
- (1,3): −ac + b² + ac = b²
- (2,1): −ab + c² + ab = c²
- (2,2): −b² + ac + ca = 2ca − b²
- (2,3): −bc + bc + a² = a²
- (3,1): −ac + ac + b² = b²
- (3,2): −bc + a² + bc = a²
- (3,3): −c² + ab + ba = 2ab − c²

So D·G is exactly the matrix whose determinant is Δ. Therefore Δ = |D|·|G|.

#### Step 3: |G| = D
Take (−1) out of R₁ of G: |G| = −|a b c; c a b; b c a|. The matrix [a b c; c a b; b c a] is D with R₂ and R₃ interchanged, so its determinant is −D. Hence |G| = −(−D) = D.

#### Conclusion
```text
Δ = |D|·|G| = D · D = D² = (a³ + b³ + c³ − 3abc)²
```

Thus |a b c; b c a; c a b|² = Δ = (a³ + b³ + c³ − 3abc)². ∎

## Check yourself

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

1. Why add C2 and C3 to C1 in D?
   A. To make C1 zero
   B. Every entry of C1 becomes a + b + c, which is taken common
   C. To swap rows
   D. To make D symmetric
   Answer: B. Each row sum is a + b + c. (Key point: Evaluating D)

2. D equals:
   A. a^3 + b^3 + c^3 - 3abc
   B. abc
   C. -(a^3 + b^3 + c^3 - 3abc)
   D. 0
   Answer: C. The expansion gives -(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca). (Key point: Value of D)

3. How is Δ linked to D?
   A. Δ is D transposed
   B. Δ is a row sum of D
   C. Δ = 2D
   D. D·G has Δ's matrix, so Δ = |D||G|
   Answer: D. Multiplying D by G gives exactly the matrix of Δ. (Key point: Product trick)

4. |G| equals:
   A. D
   B. -D
   C. 0
   D. D^2
   Answer: A. Taking -1 from R1 and swapping R2, R3 gives (-1)(-1)D = D. (Key point: |G|)

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