Question

Question 5: Multiple Choices Assume that vi,2,ig are vectors in R3. Let S span ,02,s and let A be the matrix whose columns are these vectors. Assume that 1 -1 1 0 0 0-3a +b-2c We can thus conclude that A. {6,6,6) is L1. B. The point (1,1 - 1) is in the span of (o,2,s) C. The nullity of A is 2 D. The rank of A is 3 E. B and C are both correct

Answer 1

There is no pivot in their third row. Therefore {v_1^Vector, v_2^Vector, v_3^Vector} is not linearly independent. (Because the third variable is free variable) (b) The point (1, -1, 1) is in the span of (v_1^Vector, v_2^Vector, v_3^Vector) if series of (1 0 0 -1 1 0 1 0 0 1 2(1) - (+ 1) -3(1) + (1) - 2(-1)) exists i.e. (1 0 0 -1 1 0 1 0 0 1 1 0) rho (A: b) = 2 = rho (A) system has infinitely many solutions = > The point (1, 1, -1) is in the span of (v_1^Vector, v_2^Vector, v_3^Vector) (c) the nullity of A is 2 (False) Because we have two pivot rows = > rank A = 2 and by rank nullity therefore dim rank A + dim null A = dim a = > 2 = dim Null a = 03 = > dim (Null A) = 1 (d) Rank of A is 2