Skip to main content

Determinant identity: 2(a + b + c)³

Short answer

A matrix is an array of numbers, while a determinant is a scalar computed from a square matrix. By C1 → C1+C2+C3, factor 2(a+b+c), then R2−R1 and R3−R1, the determinant becomes 2(a+b+c)·1·( a+b+c)·( a+b+c) = 2(a+b+c)³.

Distinguish a matrix and a determinant. Use the property of determinants to prove: [a+b+2c a b; c b+c+2a b; c a c+a+2b] (determinant) = 2(a + b + c)³.

Difference between a matrix and a determinant

MatrixDeterminant
An arrangement (array) of numbers in rows and columns, written in brackets [ ]A single number (value) calculated from a square array, written between bars | |
Can be of any order m × nDefined only for a square array (n × n)
Has no numerical value of its ownHas a definite value
Interchanging rows and columns gives a different matrix (the transpose)Interchanging rows and columns does not change the value
Multiplying by k multiplies every elementMultiplying by k multiplies only one row (or column)

Proof

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)

Related questions

All Mathematics III chapter 1 questions →