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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2082 Baishakh, 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-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Show, using the properties of determinants, that [1+a 1 1 1; 1 1+b 1 1; 1 1 1+c 1; 1 1 1 1+d] (determinant) = abcd(1 + 1/a + 1/b + 1/c + 1/d).

## Short answer

Take a, b, c, d common from rows; apply R1 → R1+R2+R3+R4 to obtain common factor T = 1 + 1/a + 1/b + 1/c + 1/d. Subtract column 1 from columns 2, 3, 4 to get a triangular matrix with unit diagonal. The determinant is abcd(1 + 1/a + 1/b + 1/c + 1/d).

## Answer

Let

```text
Δ = | 1+a  1    1    1  |
    |  1  1+b   1    1  |
    |  1   1   1+c   1  |
    |  1   1    1   1+d |
```

#### Step 1: Take a, b, c, d common from R₁, R₂, R₃, R₄
Row 1 divided by a is (1/a + 1, 1/a, 1/a, 1/a), and similarly for the other rows:

```text
Δ = abcd | 1+1/a   1/a     1/a     1/a   |
         |  1/b   1+1/b    1/b     1/b   |
         |  1/c    1/c    1+1/c    1/c   |
         |  1/d    1/d     1/d    1+1/d  |
```

#### Step 2: R₁ → R₁ + R₂ + R₃ + R₄
Every column has the same sum

```text
T = 1 + 1/a + 1/b + 1/c + 1/d
```

so the new first row is (T, T, T, T). Take T common:

```text
Δ = abcd · T | 1      1      1      1    |
             | 1/b  1+1/b   1/b    1/b   |
             | 1/c   1/c   1+1/c   1/c   |
             | 1/d   1/d    1/d   1+1/d  |
```

#### Step 3: C₂ → C₂ − C₁, C₃ → C₃ − C₁, C₄ → C₄ − C₁
```text
Δ = abcd · T | 1    0  0  0 |
             | 1/b  1  0  0 |
             | 1/c  0  1  0 |
             | 1/d  0  0  1 |
```

The matrix is lower triangular with diagonal 1, 1, 1, 1, so its determinant is 1.

**Δ = abcd (1 + 1/a + 1/b + 1/c + 1/d).** ∎

## Check yourself

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

1. Taking a, b, c, d from R1..R4 gives a factor
   A. abcd
   B. a+b+c+d
   C. 1
   D. abcd(1/a)
   Answer: A. Each row is divided by its own variable. (Key point: First step)

2. After R1 -> R1+R2+R3+R4 the new first row is
   A. (1,1,1,1)
   B. (T, T, T, T), T = 1 + 1/a + 1/b + 1/c + 1/d
   C. (0,0,0,0)
   D. (a,b,c,d)
   Answer: B. Each column sums to T. (Key point: Row sum)

3. Why is the remaining determinant 1?
   A. Its first row is zero
   B. It is symmetric
   C. It is triangular with diagonal 1s
   D. It equals T
   Answer: C. Column operations C_j - C_1 clear row 1. (Key point: Triangular)

## 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: 2(a+b+c)³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Adeterminant-identity-2-a-b-c) — Mathematics III, chapter 1, asked 6×

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