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Determinant identity: 4a²b²c²

Short answer

Take common factors a, b, c from columns. Then C₁→C₁+C₂−C₃, R₃→R₃−R₂, and expand along C₁ to get determinant = 4a²b²c².

Use properties of determinants to prove: [a² bc ac+c²; a²+ab b² ac; ab b²+bc c²] (determinant) = 4a²b²c².

Let

Δ = | a²     bc     ac+c² |
    | a²+ab  b²     ac    |
    | ab     b²+bc  c²    |

Step 1: Take common factors from the columns.

  • C₁ = (a², a(a+b), ab) has the factor a
  • C₂ = (bc, b², b(b+c)) has the factor b
  • C₃ = (c(a+c), ac, c²) has the factor c
Δ = abc | a    c    a+c |
        | a+b  b    a   |
        | b    b+c  c   |

Step 2: C₁ → C₁ + C₂ − C₃.

  • Row 1: a + c − (a + c) = 0
  • Row 2: (a + b) + b − a = 2b
  • Row 3: b + (b + c) − c = 2b
Δ = abc | 0   c    a+c |
        | 2b  b    a   |
        | 2b  b+c  c   |

Step 3: R₃ → R₃ − R₂.

Δ = abc | 0   c    a+c |
        | 2b  b    a   |
        | 0   c    c−a |

Step 4: Expand along C₁. The only non-zero entry is 2b at (2, 1), with sign (−1)^(2+1) = −1:

Δ = abc · (−2b) · | c  a+c |
                  | c  c−a |
  = abc · (−2b) · [ c(c−a) − c(a+c) ]
  = abc · (−2b) · [ −2ac ]
  = abc · 4abc

Δ = 4a²b²c². ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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