S1

Question 1

A system of linear equations has infinitely many solutions if and only if it has a free variable.

Answer 1

False:

Take for instance a system with the augmented matrix.

The third row says that the third variable is a free variable, but the fourth equation is 0 = 1, so actually there are no solutions.

Question 2

A vector b is a linear combination of the columns of the matrix A if and only if Ax = b is consistent.

Answer 2

True

Write A = [a₁ ···a_n]. We have a linear combination b = c₁a₁ +···+c_na_n, if and only if the vector x = (c₁,…,c_n) is a solution to Ax = b.

Question 3

If u ∈ Span {v₁,…,v_n}, then v₁ ∈ Span {u,v₂,…,v_n}.

Answer 3

False

Take for example v₁ = (1,0), v₂ = (0,1) ,and u = (0,1).

Then u ∈ Span {v₁,v₂} because u = v₂, but v₁ ∉ Span {u,v₂} = Span{v₂}, since this only consists of vectors of the form (0,a).

Question 4

For u, v ∈ ℝ³, Span{u, v} is always a plane.

Answer 4

False

Take for instance u = (1,0,0) and v = (2,0,0), then the Span {u, v} is the line consisting of all vectors of the form (a,0,0).

Question 5

If A= [a₁···a_n] is a n x m matrix and Span {a₁…a_n} ≠ ℝ^m, then there is a b ∈ ℝ^m such that Ax = b is inconsistent.

Answer 5

True

If Span {a₁ . . . a_n} ≠ ℝ^m, then, since the span is the set of all linear combinations of the given vectors, there must be some b ∈ ℝ^m which is not a linear combination of a₁ . . . a_n. That means that Ax = b has no solution, or in other words, it is an inconsistent system.