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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2071 Chaitra, 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-proof-nine
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Using the properties, evaluate the determinant [1 a a² a³+bcd; 1 b b² b³+cda; 1 c c² c³+abd; 1 d d² d³+abc].

## Short answer

Multiply rows by a, b, c, d, split last column into two determinants. The first yields abcd times Vandermonde, the second yields −abcd times Vandermonde, sum zero, so the determinant equals zero.

## Answer

Let

```text
Δ = | 1  a  a²  a³+bcd |
    | 1  b  b²  b³+cda |
    | 1  c  c²  c³+abd |
    | 1  d  d²  d³+abc |
```

**Step 1: Multiply R₁, R₂, R₃, R₄ by a, b, c, d.** The determinant becomes abcd·Δ, and the last column entries become a⁴ + abcd, b⁴ + abcd, c⁴ + abcd, d⁴ + abcd:

```text
abcd·Δ = | a  a²  a³  a⁴+abcd |
         | b  b²  b³  b⁴+abcd |
         | c  c²  c³  c⁴+abcd |
         | d  d²  d³  d⁴+abcd |
```

**Step 2: Split the last column** (additive property of determinants):

```text
abcd·Δ = | a a² a³ a⁴ |         | a a² a³ 1 |
         | b b² b³ b⁴ |  + abcd | b b² b³ 1 |
         | c c² c³ c⁴ |         | c c² c³ 1 |
         | d d² d³ d⁴ |         | d d² d³ 1 |
```

**Step 3: First determinant.** Take a, b, c, d common from the rows. It equals abcd·V, where V is the Vandermonde determinant

```text
V = | 1 a a² a³ |
    | 1 b b² b³ |  = Π (xⱼ − xᵢ) over i < j
    | 1 c c² c³ |
    | 1 d d² d³ |
```

**Step 4: Second determinant.** Move the column of ones to the front. This needs three adjacent column interchanges, so the sign is (−1)³ = −1; the determinant equals −V.

**Step 5: Combine.**

```text
abcd·Δ = abcd·V + abcd·(−V) = 0
```

Since abcd ≠ 0 in general (and both sides are polynomials in a, b, c, d), Δ = 0.

**Answer: Δ = 0.**

## Check yourself

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

1. Multiplying the rows by a, b, c, d makes the last column entries:
   A. a^3 + bcd
   B. a^4 + abcd, b^4 + abcd, etc.
   C. a^4 only
   D. abcd only
   Answer: B. a(a^3 + bcd) = a^4 + abcd, and the determinant becomes abcd·Δ. (Key point: Multiplying rows)

2. Which property splits the last column?
   A. Row swapping
   B. Scalar multiplication
   C. Additivity of the determinant in a column
   D. Transpose
   Answer: C. A column that is a sum splits into two determinants. (Key point: Splitting a column)

3. The determinant with columns a, a^2, a^3, a^4 equals:
   A. V
   B. abcd · V, V being the Vandermonde determinant
   C. 0
   D. abcd + V
   Answer: B. Taking a, b, c, d out of the rows leaves V. (Key point: Vandermonde)

4. Moving the column of ones to the front by three adjacent swaps gives sign:
   A. +1
   B. -1
   C. 0
   D. 3
   Answer: B. (-1)^3 = -1, so that determinant equals -V. (Key point: Column movement sign)

5. abcd·V + abcd·(-V) = 0, so Δ is:
   A. V
   B. abcd
   C. 1
   D. 0
   Answer: D. The two terms cancel and Δ = 0. (Key point: Result)

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