# Thread: show that the vectors are linearly dependent

1. ## show that the vectors are linearly dependent

Show that the vectors u [-1 1] v [2 1] w[3 3] are linearly dependent.

So, linearly dependent means that there is a solution for A,B,C that doesn't involve all zero's right?

So, -a + 2b + 3c = 0 and a + b + 3c = 0

I just solve these linear system of equations to show that the vectors are linearly dependent? Or am I completely wrong?

Thank you

2. Correct.

The system of equations

\displaystyle \begin{aligned} -a+2b+3c&=0\\ a+b+3c&=0 \end{aligned}

can be solved by linear combination. Hint: Multiple solutions will work.

3. Uhh, it has been a while since I solved a system of equations, especially with 3 variables.. is there a good method for doing this besides just "looking at it". I must be doing something wrong because I keep getting zero for all of the variables when I try to solve it..

4. Originally Posted by zodiacbrave
Show that the vectors u [-1 1] v [2 1] w[3 3] are linearly dependent.

So, linearly dependent means that there is a solution for A,B,C that doesn't involve all zero's right?

So, -a + 2b + 3c = 0 and a + b + 3c = 0

I just solve these linear system of equations to show that the vectors are linearly dependent? Or am I completely wrong?

Thank you

You could also note that since

$\displaystyle u,v,w \in \mathbb{R}^2$ (It's basis has exactly 2 vectors)

You have 3 vectors in a 2D space so they must be linearly dependant

5. Originally Posted by zodiacbrave
Show that the vectors u [-1 1] v [2 1] w[3 3] are linearly dependent.

So, linearly dependent means that there is a solution for A,B,C that doesn't involve all zero's right?

So, -a + 2b + 3c = 0 and a + b + 3c = 0

I just solve these linear system of equations to show that the vectors are linearly dependent? Or am I completely wrong?

Thank you
$\displaystyle u,v,w$ being linearly dependent over the field $\displaystyle F$ means that there exist scalars $\displaystyle a_1,a_2,a_3 \in F$, not all 0, such that $\displaystyle a_1v + a_2u + a_3w = 0$

6. Originally Posted by zodiacbrave
Uhh, it has been a while since I solved a system of equations, especially with 3 variables.. is there a good method for doing this besides just "looking at it". I must be doing something wrong because I keep getting zero for all of the variables when I try to solve it..
Subtracting the first equation from the second gives us our first result:

\displaystyle \begin{aligned} (a - (-a))+(b-2b)+(3c-3c)&=0\\ 2a-b&=0\\ 2a&=b. \end{aligned}

7. Right, so I get b = -2a then I substitute it back into one of the equations and find that a = c and then i put all of that into an equation and find that a = 0.. but from all of that i get a = 0, b = 0 and c = 0 which is all wrong..

8. Originally Posted by zodiacbrave
Right, so I get b = -2a then I substitute it back into one of the equations and find that a = c and then i put all of that into an equation and find that a = 0.. but from all of that i get a = 0, b = 0 and c = 0 which is all wrong..
The correct equation is $\displaystyle b=2a$. Substituting, we obtain

\displaystyle \begin{aligned} -a+2(2a)+3c&=0\\ 3a+3c&=0.\end{aligned}

9. Okay,

so if b = 2a, I subsitute it back into the equation and get 3a + 3c = 0

then if I solve this for a, I get a = -c..

No matter where I go with this, I end up with all zero answers. Is there something simple I am missing here? Maybe I need to look up system of linear equations?

Thank you