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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2069 Chaitra, 5 marks
- 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-eleven
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the value of the determinant [a² a²−(b−c)² bc; b² b²−(c−a)² ca; c² c²−(a−b)² ab].

## Short answer

Apply column operations: C₂ → C₂ − C₁, then factor (b−c)² etc., add 2C₃, then C₂ + C₁ to get common factor. Result is (a−b)(b−c)(c−a)(a+b+c)(a²+b²+c²).

## Answer

Let

```text
Δ = | a²  a²−(b−c)²  bc |
    | b²  b²−(c−a)²  ca |
    | c²  c²−(a−b)²  ab |
```

**Step 1: C₂ → C₂ − C₁.** The second column becomes −(b−c)², −(c−a)², −(a−b)². Take (−1) common:

```text
Δ = −| a²  (b−c)²  bc |
     | b²  (c−a)²  ca |
     | c²  (a−b)²  ab |
```

**Step 2: C₂ → C₂ + 2C₃.** Since (b−c)² + 2bc = b² + c², and so on:

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

**Step 3: C₂ → C₂ + C₁.** All entries of the second column become a² + b² + c²; take it common:

```text
Δ = −(a²+b²+c²) | a²  1  bc |
                | b²  1  ca |
                | c²  1  ab |
```

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

```text
= −(a²+b²+c²) | (a−b)(a+b)   0  −c(a−b) |
              | (b−c)(b+c)   0  −a(b−c) |
              |     c²       1     ab    |
```

**Step 5: Expand along C₂** (entry 1 at (3, 2), sign −1) and take (a−b) and (b−c) out:

```text
= (a²+b²+c²)(a−b)(b−c) | a+b  −c |
                        | b+c  −a |
= (a²+b²+c²)(a−b)(b−c) [ −a(a+b) + c(b+c) ]
= (a²+b²+c²)(a−b)(b−c)(c−a)(a+b+c)
```

because −a² − ab + bc + c² = (c² − a²) + b(c − a) = (c − a)(a + b + c).

**Answer: Δ = (a − b)(b − c)(c − a)(a + b + c)(a² + b² + c²).**

## Check yourself

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

1. After C2 -> C2 - C1, the second column entries are:
   A. b^2 + c^2
   B. -(b-c)^2, -(c-a)^2, -(a-b)^2
   C. 0
   D. a^2 - b^2
   Answer: B. a^2 - (b-c)^2 - a^2 = -(b-c)^2. (Key point: Step 1)

2. C2 -> C2 + 2C3 uses the identity:
   A. (b-c)^2 + 2bc = b^2 + c^2
   B. (b+c)^2 = b^2 + c^2
   C. (b-c)^2 = b^2 - c^2
   D. bc = b + c
   Answer: A. Adding 2bc converts the square of the difference into b^2 + c^2. (Key point: Step 2)

3. After C2 -> C2 + C1, the second column is all:
   A. 1
   B. a^2 + b^2 + c^2
   C. abc
   D. a + b + c
   Answer: B. Each entry becomes a^2 + b^2 + c^2, taken common. (Key point: Common factor)

4. The determinant equals:
   A. (a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)
   B. (a-b)(b-c)(c-a)(a^2+b^2+c^2)
   C. abc(a+b+c)
   D. 0
   Answer: A. The final 2x2 gives (c-a)(a+b+c). (Key point: Final answer)

## Related in Mathematics III

- [Determinant identity: 2(ab + bc + ca)³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-2-ab-bc-ca) — Mathematics III, chapter 1
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