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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2076 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-five
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Prove that [1+a²−b² 2ab −2b; 2ab 1−a²+b² 2a; 2b −2a 1−a²−b²] (determinant) = (1 + a² + b²)³ by using the properties of determinants.

## Short answer

Apply R₁ → R₁ + bR₃ and R₂ → R₂ − aR₃, then factor k = 1+a²+b² from the first two rows. Expanding the simplified 3×3 determinant gives k, making the original determinant k³ = (1+a²+b²)³.

## Answer

Let k = 1 + a² + b² and

```text
Δ = | 1+a²−b²    2ab       −2b     |
    |  2ab     1−a²+b²      2a     |
    |  2b       −2a      1−a²−b²   |
```

**Step 1: R₁ → R₁ + b·R₃ and R₂ → R₂ − a·R₃.**

Row 1:
- (1 + a² − b²) + 2b² = 1 + a² + b² = k
- 2ab + b(−2a) = 0
- −2b + b(1 − a² − b²) = −b − a²b − b³ = −bk

Row 2:
- 2ab − a(2b) = 0
- (1 − a² + b²) − a(−2a) = 1 + a² + b² = k
- 2a − a(1 − a² − b²) = a + a³ + ab² = ak

```text
Δ = | k    0    −bk |
    | 0    k     ak |
    | 2b  −2a  1−a²−b² |
```

**Step 2:** Take k common from R₁ and from R₂ (factor k²):

```text
Δ = k² | 1    0   −b |
       | 0    1    a |
       | 2b  −2a  1−a²−b² |
```

**Step 3:** Expand along R₁:

```text
= 1·[1·(1−a²−b²) − a·(−2a)] − 0 + (−b)·[0·(−2a) − 1·2b]
= (1 − a² − b² + 2a²) + 2b²
= 1 + a² + b² = k
```

Therefore Δ = k² · k = k³ = **(1 + a² + b²)³.** ∎

## Check yourself

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

1. Which operations are used first?
   A. R1 -> R1 + bR3 and R2 -> R2 - aR3
   B. R1 -> R1 - R2 only
   C. C1 -> C1 + C2 + C3
   D. R3 -> R3 + R1 + R2
   Answer: A. These produce k = 1 + a^2 + b^2 in the diagonal places of rows 1 and 2. (Key point: Row operations)

2. After the operation, what is the first entry of row 1?
   A. 1 + a^2 - b^2
   B. k = 1 + a^2 + b^2
   C. 0
   D. -bk
   Answer: B. (1 + a^2 - b^2) + b(2b) = 1 + a^2 + b^2. (Key point: Row 1 result)

3. What is taken out of rows 1 and 2?
   A. k in total
   B. a and b
   C. k from each, giving k^2
   D. 2k
   Answer: C. Both rows have k common, giving k^2 times a simpler determinant. (Key point: Taking common factors)

4. The remaining 3x3 determinant evaluates to k, so the answer is:
   A. k^2
   B. k^3
   C. 2k^2
   D. k^4
   Answer: B. k^2 · k = (1 + a^2 + b^2)^3. (Key point: Final value)

## Close variants set in other exams

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