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Determinant identity proof

Short answer

Take a, b, c, d common from rows; apply R1 → R1+R2+R3+R4 to obtain common factor T = 1 + 1/a + 1/b + 1/c + 1/d. Subtract column 1 from columns 2, 3, 4 to get a triangular matrix with unit diagonal. The determinant is abcd(1 + 1/a + 1/b + 1/c + 1/d).

Show, using the properties of determinants, that [1+a 1 1 1; 1 1+b 1 1; 1 1 1+c 1; 1 1 1 1+d] (determinant) = abcd(1 + 1/a + 1/b + 1/c + 1/d).

Let

Δ = | 1+a  1    1    1  |
    |  1  1+b   1    1  |
    |  1   1   1+c   1  |
    |  1   1    1   1+d |

Step 1: Take a, b, c, d common from R₁, R₂, R₃, R₄

Row 1 divided by a is (1/a + 1, 1/a, 1/a, 1/a), and similarly for the other rows:

Δ = abcd | 1+1/a   1/a     1/a     1/a   |
         |  1/b   1+1/b    1/b     1/b   |
         |  1/c    1/c    1+1/c    1/c   |
         |  1/d    1/d     1/d    1+1/d  |

Step 2: R₁ → R₁ + R₂ + R₃ + R₄

Every column has the same sum

T = 1 + 1/a + 1/b + 1/c + 1/d

so the new first row is (T, T, T, T). Take T common:

Δ = abcd · T | 1      1      1      1    |
             | 1/b  1+1/b   1/b    1/b   |
             | 1/c   1/c   1+1/c   1/c   |
             | 1/d   1/d    1/d   1+1/d  |

Step 3: C₂ → C₂ − C₁, C₃ → C₃ − C₁, C₄ → C₄ − C₁

Δ = abcd · T | 1    0  0  0 |
             | 1/b  1  0  0 |
             | 1/c  0  1  0 |
             | 1/d  0  0  1 |

The matrix is lower triangular with diagonal 1, 1, 1, 1, so its determinant is 1.

Δ = abcd (1 + 1/a + 1/b + 1/c + 1/d). ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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