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

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2069 Asar, 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-ten
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the value of the determinant [1 a a² a³+bcd; 1 b b² b³+cda; 1 c c² c³+dab; 1 d d² d³+abc].

## Short answer

After multiplying rows by a,b,c,d and splitting, the two determinants cancel. The result is 0.

## Answer

Evaluate

```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 R1..R4 by a, b, c, d turns the last column into:
   A. a^3 + bcd, ...
   B. a^4 + abcd, b^4 + abcd, c^4 + abcd, d^4 + abcd
   C. abcd, abcd, abcd, abcd
   D. a^4, b^4, c^4, d^4
   Answer: B. Each entry becomes the fourth power plus abcd. (Key point: Row multiplication)

2. Splitting the last column gives two determinants:
   A. One with a^4, ... and abcd times one with a column of ones
   B. Two with a^4
   C. One zero and one V
   D. Both with ones
   Answer: A. Additivity gives the sum of the power-column determinant and abcd times the ones-column determinant. (Key point: Splitting)

3. The ones-column determinant equals -V because:
   A. It has a zero row
   B. Three adjacent column interchanges are needed, each changing sign
   C. V is negative
   D. A column is doubled
   Answer: B. Moving the ones column to the front takes 3 swaps, (-1)^3 = -1. (Key point: Sign)

4. Hence Δ equals:
   A. V
   B. abcd
   C. 2V
   D. 0
   Answer: D. abcd·Δ = abcd·V - abcd·V = 0. (Key point: Answer)

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