---
title: "Determinant identity: 2(a+b+c)³"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:determinant-identity-2-a-b-c"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-2-a-b-c
site: BCT Engineering
---

# Determinant identity: 2(a+b+c)³

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2083 Baishakh, 5 marks
- Video: [Show that: det [[a - b - c, 2a, 2a], [2b, b - c - a, 2b], [2c, 2c, c - a - b]] = (a + b + c)^3](https://www.youtube.com/watch?v=cwJx3yqzVAA) — Mathematics Untold
- 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-2-a-b-c
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Using properties of determinants, show that [a+b+2c a b; c b+c+2a b; c a c+a+2b] (determinant) = 2(a+b+c)³.

## Short answer

Let s = a+b+c. Perform C₁ → C₁ + C₂ + C₃, factor 2s, then R₂ → R₂ − R₁, R₃ → R₃ − R₁ to get an upper triangular matrix with diagonal entries 1, s, s. The determinant is 2s · s² = 2(a+b+c)³.

## Answer

Let s = a + b + c and

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

**Step 1: C₁ → C₁ + C₂ + C₃.** Each entry of the new first column is 2(a + b + c):

- Row 1: (a + b + 2c) + a + b = 2a + 2b + 2c
- Row 2: c + (b + c + 2a) + b = 2a + 2b + 2c
- Row 3: c + a + (c + a + 2b) = 2a + 2b + 2c

```text
Δ = | 2s      a        b    |
    | 2s   b+c+2a      b    |
    | 2s      a      c+a+2b |
```

**Step 2:** Take 2s common from C₁:

```text
Δ = 2s | 1      a        b    |
       | 1   b+c+2a      b    |
       | 1      a      c+a+2b |
```

**Step 3: R₂ → R₂ − R₁, R₃ → R₃ − R₁.**

- R₂ − R₁ = (0, b + c + 2a − a, 0) = (0, s, 0)
- R₃ − R₁ = (0, 0, c + a + 2b − b) = (0, 0, s)

```text
Δ = 2s | 1  a  b |
       | 0  s  0 |
       | 0  0  s |
```

This is upper triangular, so the determinant is the product of the diagonal: 1 · s · s = s².

Δ = 2s · s² = **2(a + b + c)³.** ∎

## Check yourself

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

1. Which column operation is used first?
   A. C1 -> C1 + C2 + C3
   B. C1 -> C1 - C2
   C. R1 -> R1 + R2
   D. C2 -> C2 - C3
   Answer: A. Each new C1 entry is 2(a+b+c). (Key point: First operation)

2. What is taken from C1?
   A. a+b+c only
   B. 2(a+b+c)
   C. abc
   D. 2
   Answer: B. Entries are all 2s. (Key point: Common factor)

3. After R2->R2-R1 and R3->R3-R1 the determinant is
   A. diagonal 1, a, b
   B. zero
   C. triangular with diagonal 1, s, s
   D. diagonal s, s, s only
   Answer: C. Product is s^2. (Key point: Triangular form)

4. The value is
   A. (a+b+c)^3
   B. 2(a+b+c)^2
   C. 4(a+b+c)^3
   D. 2(a+b+c)^3
   Answer: D. 2s times s^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) — Mathematics III, chapter 1, asked 6×
- [Determinant identity proof](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-proof-two) — Mathematics III, chapter 1, asked 4×
- [Determinant identity: (a + b + c)³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-a-b-c-3) — Mathematics III, chapter 1, asked 6×
- [Determinant identity proof](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-proof-seven) — Mathematics III, chapter 1, asked 6×

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