Linear Algebra

310 Chapter Four. Determinants3.10 Definition The permutation expansion for determinants is∣ t 1,1 t 1,2 . . . t ∣∣∣∣∣∣∣∣∣ 1,nt 2,1 t 2,2 . . . t 2,n= t 1,φ1 (1)t 2,φ1 (2) · · · t n,φ1 (n)|P φ1 |.+ t∣1,φ2 (1)t 2,φ2 (2) · · · t n,φ2 (n)|P φ2 |t n,1 t n,2 . . . t n,n.+ t 1,φk (1)t 2,φk (2) · · · t n,φk (n)|P φk |where φ 1 , . . . , φ k are all of the n-permutations.This formula is often written in summation notation∑|T| = t 1,φ(1) t 2,φ(2) · · · t n,φ(n) |P φ |permutations φread aloud as, “the sum, over all permutations φ, of terms having the formt 1,φ(1) t 2,φ(2) · · · t n,φ(n) |P φ |.”3.11 Example The familiar 2×2 determinant formula follows from the above.∣ t 1,1 t ∣∣∣∣1,2= t∣1,1 t 2,2 · |P φ1 | + t 1,2 t 2,1 · |P φ2 |t 2,1 t 2,2 = t 1,1 t 2,2 ·1 0∣0 1∣ + t 0 11,2t 2,1 ·∣1 0∣= t 1,1 t 2,2 − t 1,2 t 2,1So does the 3×3 determinant formula.∣ t 1,1 t 1,2 t ∣∣∣∣∣∣1,3t 2,1 t 2,2 t 2,3 = t 1,1 t 2,2 t 3,3 |P φ1 | + t 1,1 t 2,3 t 3,2 |P φ2 | + t 1,2 t 2,1 t 3,3 |P φ3 |∣t 3,1 t 3,2 t 3,3 + t 1,2 t 2,3 t 3,1 |P φ4 | + t 1,3 t 2,1 t 3,2 |P φ5 | + t 1,3 t 2,2 t 3,1 |P φ6 |= t 1,1 t 2,2 t 3,3 − t 1,1 t 2,3 t 3,2 − t 1,2 t 2,1 t 3,3+ t 1,2 t 2,3 t 3,1 + t 1,3 t 2,1 t 3,2 − t 1,3 t 2,2 t 3,1Computing a determinant by permutation expansion usually takes longerthan Gauss’s Method. However, while it is not often used in practice, we use itfor the theory, to prove that the determinant function is well-defined.We will just state the result here and defer its proof to the following subsection.3.12 Theorem For each n there is an n×n determinant function.Also in the next subsection is the proof of this result (these two proofs sharesome features).