The answers are TFFFFT but I dont understand why Linear Alge

The answers are T,F,F,F,F,T but I don\'t understand why. Linear Algebra and its Applications.

4. Suppose u1,u2,u3,b are in Rm, A = [u1 u2 u3], and x is in R3.

True or False. Justify answer/show work.

a) If Ax = b has a solution for every b and m >= 3, then {u1, u2, u3} is linearly independent.

b) {u1, u2, u3} is linearly independent if u3 = 2u1 + 3u2.

c) If Ax = b has a solution for every b, then m>3.

d) If Ax = 0 has only the trivial solution, then for every b Ax = b has exactly one solution.

e) {u1, u2, u3} is linearly independent, then {u1,u2} is linearly dependent.

f) If {u1, u2} is linearly independent and {u1, u2, u3} is linearly dependent, then u3 span is in Span {u1, u2}.

Solution

a)

b lies in column space of A ie span{u1,u2,u3}

So we have span{u1,u2,u3}=R3

HEnce, u1,u2,u3 form basis for R3 as R3 has dimension 3 so they must be linearly independent

b)

False

u3=2u1+3u2

means

2u1+3u2+(-1)*u3=0

Hence they are linearly dependent

c)

False

Ax=b has solution for every b means

any b in Rm is in span of {u1,u2,u3}

ie span{u1,u2,u3}=Rm

Hence, dim(Rm)=m<=3

d)

Rank(A)+nullity(A)=3

IF Ax=0 has only trivial solution then nullity =0

So, rank(A)=3

So u1,u2,u3 are linearly independent

So each equation Ax=b has unique solution

e)

False.

u1,u2,u3 must be linearly independent pairwise also for all three to be linearly indepdnent

f)

True.

SInec, u1,u2,u3 are linearly dependent so there is some, a,b,c so that

au1+bu2+cu3=0 not all a,b,c being equal to 0

Case 1: c=0

So, au1+bu2=0

So one of a and b must be non zero which is not possible as u1,u2 are linearly independent

So, a=b=0

which is contradiction

So we must have:c non zero

So we can write

u3=-(a/c)u1-(b/c)u2