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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2075 Asoj, 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-six
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Define the determinant as a function and, using its properties, show that [b+c c+a a+b; q+r r+p p+q; y+z z+x x+y] (determinant) = 2 [a p x; b q y; c r z] (determinant).

## Short answer

A determinant is an alternating multilinear function assigning to each square matrix a number via permutations. Apply C₁ → C₁+C₂+C₃, then C₂−C₁, C₃−C₁, and take factors; the result equals twice the determinant of the transposed matrix.

## Answer

#### Determinant as a function
For a square matrix A = [aᵢⱼ] of order n, the determinant is a function det : Mₙ → ℝ (or ℂ) that assigns to A the single number

```text
|A| = Σ sgn(π) · a₁π(1) · a₂π(2) · … · aₙπ(n)
```

summed over all n! permutations π of {1, …, n}, with sgn(π) = ±1 according as π is even or odd. This function is linear in each row (or column) separately, changes sign when two rows are interchanged, and equals 1 for the identity matrix. These properties are used below.

#### Proof
Let

```text
Δ = | b+c  c+a  a+b |
    | q+r  r+p  p+q |
    | y+z  z+x  x+y |
```

**Step 1: C₁ → C₁ + C₂ + C₃.** Each entry of the new first column is twice a sum:

```text
Δ = | 2(a+b+c)  c+a  a+b |
    | 2(p+q+r)  r+p  p+q |      = 2 | a+b+c  c+a  a+b |
    | 2(x+y+z)  z+x  x+y |          | p+q+r  r+p  p+q |
                                    | x+y+z  z+x  x+y |
```

**Step 2: C₂ → C₂ − C₁, C₃ → C₃ − C₁.** Then c + a − (a+b+c) = −b and a + b − (a+b+c) = −c, and similarly for the other rows:

```text
Δ = 2 | a+b+c  −b  −c |
      | p+q+r  −q  −r |
      | x+y+z  −y  −z |
```

**Step 3: C₁ → C₁ + C₂ + C₃.**

```text
Δ = 2 | a  −b  −c |
      | p  −q  −r |
      | x  −y  −z |
```

**Step 4:** Take (−1) out of C₂ and of C₃; the factor is (−1)(−1) = 1:

```text
Δ = 2 | a  b  c |
      | p  q  r |
      | x  y  z |
```

Finally, a determinant is unchanged when rows and columns are interchanged (|M| = |Mᵀ|):

**Δ = 2 | a p x ; b q y ; c r z |.** ∎

## Check yourself

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

1. Why does C1 -> C1 + C2 + C3 give a factor 2?
   A. Each row sum is twice (a+b+c)-type sum, e.g. (b+c)+(c+a)+(a+b) = 2(a+b+c)
   B. Because the determinant is doubled by any column operation
   C. Because C2 = C3
   D. Because the matrix is symmetric
   Answer: A. Each new first-column entry is twice a sum, so 2 is taken out. (Key point: Step 1)

2. After C2 -> C2 - C1, the entry c + a - (a+b+c) becomes:
   A. b
   B. -b
   C. c
   D. 0
   Answer: B. c + a - a - b - c = -b. (Key point: Step 2)

3. Taking (-1) out of C2 and C3 gives a total factor of:
   A. -1
   B. 0
   C. 2
   D. 1
   Answer: D. (-1)(-1) = 1. (Key point: Sign factors)

4. Which property turns the rows (a b c; p q r; x y z) into the required form?
   A. Swapping two rows
   B. Multiplying a row by a scalar
   C. |M| = |M^T|
   D. Adding rows
   Answer: C. The determinant is unchanged when rows and columns are interchanged. (Key point: Transpose property)

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