1. ## [SOLVED] linearly independent?

How do I determine whether or not the following vectors are linearly independent?

x (3, 1, 0, 1) y (2, 0, 1, -1) z (6, 2, 1, -2)

2. $v_1, v_2, v_3$ are lin.indep, if from $\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \alpha_3\cdot v_3 = 0$
$\Rightarrow \alpha_1=\alpha_2=\alpha_3=0$

Than, let see the system:
$\alpha_1(-3, 1, 0, 1) + \alpha_2 (2, 0, 1, 1) + \alpha_3(6, 2, 1, 2)= 0$

$-3\alpha_1 + 2\alpha_2 + 6\alpha_3 =0$
$1\alpha_1 + 0\alpha_2 + 2\alpha_3 =0$
$0\alpha_1 + 1\alpha_2 + 1\alpha_3 =0$
$1\alpha_1 + 1\alpha_2 + 2\alpha_3 =0$

Solve it...

If you get $\alpha_1=\alpha_2=\alpha_3=0$ than vectors is lineary indipendent, else not.

3. Originally Posted by pungakitty
How do I determine whether or not the following vectors in R4 is linearly independent?

(-3, 1, 0, 1) (2, 0, 1, 1) (6, 2, 1, 2)
Another method is to place these as column vectors into a 4 x 3 matrix and put them in echelon form. If the rank is 3, then the are independent. If it's 2 or less, they are dependent.

This is if you prefer matrix operations over solving linear equations.

4. I don't think I'm doing it right. I put it in matrix form and came up with

[1 0 2]
[0 1 6]
[0 0 -5]
[0 0 -6]

That can't be an answer, can it? I am so lost.

5. Originally Posted by pungakitty
I don't think I'm doing it right. I put it in matrix form and came up with

[1 0 2]
[0 1 6]
[0 0 -5]
[0 0 -6]

That can't be an answer, can it? I am so lost.
Your are not completely done in transforming your matrix to row-reduced echelon form.

As GnomeSain mentioned in the above post, you should place the given cloumn vectors into a 4x3 matrix, and its row reduced echelon form looks like this:

$\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\\0&0&0 \end{array}\right)$

Now, can you tell, what is the rank of the matrix?

Again look at GnomeSain's post, to know how you would be able to determine linear independence by knowing the rank of the row reduced matrix!