1. ## Linear independence

v=(6, -3, 0)
u=(-3, 5, -7)
w=(9, 6, k)

what real number can't k be for the vectors to be linearly independent?

2. Hi

What about calculating the determinant of the three vectors ? When it's null, the vectors are linearly dependent and it'll give you the condition on $k$.

3. Originally Posted by weasley74
v=(6, -3, 0)
u=(-3, 5, -7)
w=(9, 6, k)

what real number can't k be for the vectors to be linearly independent?
set up the augmented matrix: $\left( \begin{array}{ccc} \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] & \left[ \begin{array}{c} u_1 \\ u_2 \\ u_3 \end{array} \right] & \left[ \begin{array}{c} w_1 \\ w_2 \\ w_3 \end{array} \right] \end{array} \right)$. equate it to zero and solve the system. the k for which the system only has the trivial solution is linearly independent

4. I've tried all of this, and I keep getting the wrong answer.

5. Then write your answers here so that we can correct them.

6. I get that k can't be 21 for the vectors to be linearly independent, please correct me

7. Maybe should I have asked for the calculations instead of the answers ? (if you get k=21 check the signs, it might help)

8. thanks! sign problem. jeez, i've been going crazy over this