Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vectors (2, 6, 6) and (−9, −2, 5) in Span{u, v} ?

Let [tex]b=(b_1,b_2,b_3) \in \mathbb{R}^3[/tex]. We have that [tex]b\in \text{Span}\{u,v\}[/tex] if and only if we can find scalars [tex] \alpha,\beta \in \mathbb{R}[/tex] such that [tex]\alpha u + \beta v = b[/tex]. This can be translated to the following equations:

1. [tex]-\alpha + 3 \beta = b_1 [/tex]

2.[tex] 2\alpha+4 \beta = b_2[/tex]

3. [tex] 3 \alpha +2 \beta = b_3[/tex]

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for [tex]\alpha,\beta[/tex] and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

[tex]-\alpha + 3 \beta = 2 [/tex]

[tex] 2\alpha+4 \beta = 6[/tex]

whose unique solutions are [tex]\alpha =1 = \beta[/tex], but note that for this values, the third equation doesn't hold (3+2 = 5 [tex]\neq [/tex] 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

[tex]-\alpha + 3 \beta = -9 [/tex]

[tex] 2\alpha+4 \beta = -2[/tex]

whose unique solutions are [tex]\alpha=3, \beta=-2[/tex]. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.