(1 | 3 | 1 2 | 2 | 2 3 | 1 | 3)

(1 | 3 | 1 2 | 2 | 2 3 | 1 | 3) = S.J.S^(-1) where S = 1/2(-2 | 1 | 10 0 | -1 | 12 2 | 0 | 14) J = (0 | 1 | 0 0 | 0 | 0 0 | 0 | 6) S^(-1) = 1/18(-7 | -7 | 11 12 | -24 | 12 1 | 1 | 1)

3 (rows) × 3 (columns)

(1 | 2 | 3 3 | 2 | 1 1 | 2 | 3)

6

0

2

1

6 λ^2 - λ^3

λ_1 = 6

λ_2 = 0

v_1 = (5, 6, 7)

v_2 = (-1, 0, 1)

∞