AB− BA = A12B21 − A21B12 A11B12 + A12B22 − A12B11
... for all α ∈ Rn , and this vector is unique since α − β = α in Rn implies β = α − α = 0. 3(d) This axiom also holds. We define −α = α for all α ∈ Rn . Then α ⊕ (−α) = α − α = 0. Uniqueness easily follows from uniqueness of additive inverses in Rn (with usual operations). 4(a) This axiom fails, as 1 · ...
... for all α ∈ Rn , and this vector is unique since α − β = α in Rn implies β = α − α = 0. 3(d) This axiom also holds. We define −α = α for all α ∈ Rn . Then α ⊕ (−α) = α − α = 0. Uniqueness easily follows from uniqueness of additive inverses in Rn (with usual operations). 4(a) This axiom fails, as 1 · ...
VECTORS C4 Worksheet C
... Find the value of the constants a and b such that line r = 3i − 5j + k + λ(2i + aj + bk) a passes through the point (9, −2, −8), b is parallel to the line r = 4j − 2k + µ(8i − 4j + 2k). ...
... Find the value of the constants a and b such that line r = 3i − 5j + k + λ(2i + aj + bk) a passes through the point (9, −2, −8), b is parallel to the line r = 4j − 2k + µ(8i − 4j + 2k). ...
