---
title: "Determinant identity: 4a²b²c²"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:determinant-identity-4a2b2c2"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-4a2b2c2
site: BCT Engineering
---

# Determinant identity: 4a²b²c²

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2073 Shrawan, 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-4a2b2c2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Use properties of determinants to prove: [a² bc ac+c²; a²+ab b² ac; ab b²+bc c²] (determinant) = 4a²b²c².

## Short answer

Take common factors a, b, c from columns. Then C₁→C₁+C₂−C₃, R₃→R₃−R₂, and expand along C₁ to get determinant = 4a²b²c².

## Answer

Let

```text
Δ = | a²     bc     ac+c² |
    | a²+ab  b²     ac    |
    | ab     b²+bc  c²    |
```

**Step 1: Take common factors from the columns.**
- C₁ = (a², a(a+b), ab) has the factor a
- C₂ = (bc, b², b(b+c)) has the factor b
- C₃ = (c(a+c), ac, c²) has the factor c

```text
Δ = abc | a    c    a+c |
        | a+b  b    a   |
        | b    b+c  c   |
```

**Step 2: C₁ → C₁ + C₂ − C₃.**
- Row 1: a + c − (a + c) = 0
- Row 2: (a + b) + b − a = 2b
- Row 3: b + (b + c) − c = 2b

```text
Δ = abc | 0   c    a+c |
        | 2b  b    a   |
        | 2b  b+c  c   |
```

**Step 3: R₃ → R₃ − R₂.**

```text
Δ = abc | 0   c    a+c |
        | 2b  b    a   |
        | 0   c    c−a |
```

**Step 4: Expand along C₁.** The only non-zero entry is 2b at (2, 1), with sign (−1)^(2+1) = −1:

```text
Δ = abc · (−2b) · | c  a+c |
                  | c  c−a |
  = abc · (−2b) · [ c(c−a) − c(a+c) ]
  = abc · (−2b) · [ −2ac ]
  = abc · 4abc
```

**Δ = 4a²b²c².** ∎

## Check yourself

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

1. Which factors are taken from C1, C2, C3 first?
   A. a, b, c
   B. a^2, b^2, c^2
   C. a+b, b+c, c+a
   D. 1, 1, 1
   Answer: A. C1 has factor a, C2 has b, C3 has c, giving abc. (Key point: Common factors)

2. C1 -> C1 + C2 - C3 makes the first entry:
   A. 2b
   B. a + c
   C. 0
   D. a
   Answer: C. a + c - (a + c) = 0, and the other entries become 2b and 2b. (Key point: Creating zeros)

3. For the entry 2b at position (2,1), the cofactor sign is:
   A. +
   B. -
   C. 0
   D. Depends on b
   Answer: B. (-1)^(2+1) = -1. (Key point: Expansion sign)

4. The remaining 2x2 determinant is -2ac, so Δ equals:
   A. 2a^2b^2c^2
   B. abc · 2abc
   C. abc · 4abc = 4a^2b^2c^2
   D. -4a^2b^2c^2
   Answer: C. abc · (-2b) · (-2ac) = 4a^2b^2c^2. (Key point: Result)

## Close variants set in other exams

- [Determinant identity proof](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-proof-eight) — Mathematics III, chapter 1, asked 6×
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