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Determinant identity: 2(a+b+c)³

Short answer

Let s = a+b+c. Perform C₁ → C₁ + C₂ + C₃, factor 2s, then R₂ → R₂ − R₁, R₃ → R₃ − R₁ to get an upper triangular matrix with diagonal entries 1, s, s. The determinant is 2s · s² = 2(a+b+c)³.

Using properties of determinants, show that [a+b+2c a b; c b+c+2a b; c a c+a+2b] (determinant) = 2(a+b+c)³.

Let s = a + b + c and

Δ = | a+b+2c      a        b    |
    |   c       b+c+2a     b    |
    |   c         a      c+a+2b |

Step 1: C₁ → C₁ + C₂ + C₃. Each entry of the new first column is 2(a + b + c):

  • Row 1: (a + b + 2c) + a + b = 2a + 2b + 2c
  • Row 2: c + (b + c + 2a) + b = 2a + 2b + 2c
  • Row 3: c + a + (c + a + 2b) = 2a + 2b + 2c
Δ = | 2s      a        b    |
    | 2s   b+c+2a      b    |
    | 2s      a      c+a+2b |

Step 2: Take 2s common from C₁:

Δ = 2s | 1      a        b    |
       | 1   b+c+2a      b    |
       | 1      a      c+a+2b |

Step 3: R₂ → R₂ − R₁, R₃ → R₃ − R₁.

  • R₂ − R₁ = (0, b + c + 2a − a, 0) = (0, s, 0)
  • R₃ − R₁ = (0, 0, c + a + 2b − b) = (0, 0, s)
Δ = 2s | 1  a  b |
       | 0  s  0 |
       | 0  0  s |

This is upper triangular, so the determinant is the product of the diagonal: 1 · s · s = s².

Δ = 2s · s² = 2(a + b + c)³. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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