MHB Linear Independence: Solving u,v,w Vector Questions

[Yankel]

Hello

I need some help solving the next question:

u,v,w are linearly independent vectors in a vector space V.

the vectors:

2u+4v+aw
u+2v
2u+bv

are linearly independent when:

1. a is not 0
2. b is not 4
3. a is not 0 OR b is not 4
4. a is not 0, and every value of b
5. a is not 0 AND b is not 4

first of all I noticed that answers 1 and 4 are the same.

I also know that x*v+y*u+z*w=0 implies x=y=z=0

but what next ?

 

[jakncoke1]

what do you mean by b not equal to 4, b is a vector no?

 

[Fernando Revilla]

2u+4v+aw
u+2v
2u+b

According to the possible solutions, the third vector should be $2u+bv$. We have:
$$\lambda_1(2u+4v+aw)+\lambda_2(u+2v)+\lambda_3(2u+bv)=0\Leftrightarrow\\(2\lambda_1+\lambda_2+2 \lambda_3)u+(4\lambda_1+2\lambda_2+b\lambda_3)v+(a\lambda_1)w=0\qquad (1)$$
By hyphothesis, $u,v,w$ are linearly independent so, $(1)$ is satisfied if and only if:
$$\left\{\begin{matrix}2\lambda_1+\lambda_2+2 \lambda_3=0\\4\lambda_1+2\lambda_2+b\lambda_3=0\\a\lambda_1=0\end{matrix}\right.\qquad (2)$$
The homogeneous linear system $(2)$ has only the trivial solution if and only if:
$$\mbox{rank }\begin{bmatrix}{2}&1&{2}\\{4}&{2}&{b}\\{a}&{0}&{0}\end{bmatrix}=3\Leftrightarrow\begin{vmatrix}{2}&1&{2}\\{4}&{2}&{b}\\{a}&{0}&{0}\end{vmatrix}\neq 0\Leftrightarrow a(b-4)\neq 0\Leftrightarrow a\neq 0\wedge b\neq 4$$
The correct answer is 5., $a$ is not $0$ and $b$ is not $4$.