---
title: "Determinant identity proof"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:determinant-identity-proof-13"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/determinant-identity-proof-13
site: BCT Engineering
---

# Determinant identity proof

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2068 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-13
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Using the properties of determinants prove that [x 1 y 1; 1 y 1 x; 1 x 1 y; y 1 x 1] (determinant) = (x + y + 2)(x − y)²(x + y − 2).

## Short answer

By row operations R1 → R1+R4, R2 → R2+R3, then subtracting half of new rows from R4 and R3, the determinant reduces to (x−y)^2((x+y)^2−4) = (x+y+2)(x−y)^2(x+y−2).

## Answer

Let

```text
Δ = | x 1 y 1 |
    | 1 y 1 x |
    | 1 x 1 y |
    | y 1 x 1 |
```
and s = x + y.

**Step 1: R₁ → R₁ + R₄ and R₂ → R₂ + R₃.**

- R₁ + R₄ = (x+y, 2, x+y, 2) = (s, 2, s, 2)
- R₂ + R₃ = (2, x+y, 2, x+y) = (2, s, 2, s)

**Step 2: R₄ → R₄ − ½R₁ and R₃ → R₃ − ½R₂ (using the new R₁, R₂).**

- R₄ − ½(R₁+R₄) = ½(R₄ − R₁) = ½(y−x, 0, x−y, 0) = ((y−x)/2)(1, 0, −1, 0)
- R₃ − ½(R₂+R₃) = ½(R₃ − R₂) = ½(0, x−y, 0, y−x) = ((x−y)/2)(0, 1, 0, −1)

These operations do not change the value of the determinant. Take the constants (x−y)/2 and (y−x)/2 out of R₃ and R₄; their product is −(x−y)²/4:

```text
Δ = −((x−y)²/4) | s  2  s  2 |
                 | 2  s  2  s |
                 | 0  1  0 −1 |
                 | 1  0 −1  0 |
```

**Step 3: C₃ → C₃ + C₁, C₄ → C₄ + C₂.**

```text
Δ = −((x−y)²/4) | s  2  2s  4  |
                 | 2  s  4   2s |
                 | 0  1  0   0  |
                 | 1  0  0   0  |
```

**Step 4: Evaluate.** Expand along R₄: its only non-zero entry is 1 at (4, 1), with sign (−1)⁵ = −1. The remaining 3 × 3 minor is [2 2s 4; s 4 2s; 1 0 0]; expand it along its last row (entry 1 at (3, 1), sign +):

```text
det = (−1) · | 2s  4 | = −(4s² − 16) = −4(s² − 4)
             | 4   2s|
```

Therefore

```text
Δ = −((x−y)²/4) · (−4)(s² − 4) = (x−y)² (s² − 4) = (x−y)² (s − 2)(s + 2)
```

**Δ = (x + y + 2)(x − y)²(x + y − 2).** ∎

(Check: x = 3, y = 1 gives rows (3,1,1,1), (1,1,1,3), (1,3,1,1), (1,1,3,1), whose determinant is 48 = 6 · 2² · 2 ✓.)

## Check yourself

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

1. What do R₁ → R₁+R₄ and R₂ → R₂+R₃ produce?
   A. Rows (x,1,x,1) and (1,y,1,y)
   B. Rows (s,2,s,2) and (2,s,2,s) with s = x+y
   C. Rows of zeros
   D. Rows (s,s,s,s) and (2,2,2,2)
   Answer: B. Adding gives (x+y,2,x+y,2) and (2,x+y,2,x+y). (Key point: First operation)

2. Why does R₄ → R₄ − ½R₁ not change the determinant?
   A. Adding a multiple of one row to another leaves the value unchanged
   B. It multiplies the determinant by ½
   C. It swaps two rows
   D. It makes a row zero
   Answer: A. Adding a multiple of another row is value-preserving. (Key point: Value change)

3. After the row operations, the constant taken out of R₃ and R₄ is:
   A. (x+y)²/4
   B. 4(x−y)
   C. (x−y)/2 only once
   D. −(x−y)²/4 combined
   Answer: D. The constants (x−y)/2 and (y−x)/2 multiply to −(x−y)²/4. (Key point: Factor from rows)

4. Final factorisation:
   A. (x+y)²(x−y)²
   B. (x−y)⁴
   C. (x+y+2)(x−y)²(x+y−2)
   D. (x+y+2)²(x−y)
   Answer: C. Δ = (x−y)²(s²−4) = (x+y+2)(x−y)²(x+y−2). (Key point: Final result)

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