Question

for which value of k will the vector u=(1;-2;k) in R^3 be a linear combination of the vectors v=(3;0;-2) and w=(2;-1;-5)

Answer 1

3 2 1
0 -1 * (c1,c2) = -2
-2 -5 k

Answer 2

pls use matrix representation

Answer 3

augment the left with the right into rref

Answer 4

Just set it as a matrix right and row reduce it

Answer 5

3 2 1
0 -1 -2
-2 -5 k

yes

Answer 6

another method is to determine for what values of k the determinant is zero

Answer 7

-(2+30+0)
3 2 1 3 2
0 -1 -2 0 -1
-2 -5 k -2 -5
+(-3k+8+0)

Answer 8

-3k+8-32 = 0
-3k = 24
k = -24/3 = -8

Answer 9

I would just rref on the 3 x 3 matrix
to get
3 0 -2
0 -1 -11/3
0 0 24/3 + k

then say the last row must be all zeros. this means k= -8

Answer 10

3 2 1
0 -1 -2
-2 -5 k

1 2/3 1/3
0 1 2
0 -11/3 (3k+2)/3

1 2/3 1/3
0 1 2
0 1 -(3k+2)/11


1 2/3 1/3
0 1 2
0 0 -(3k+2)-22/11

as long as the last row is zeros, this thing is "dependant"

-(3k+2)-22/11 = 0

-3k-2-22 = 0

-3k-24 = 0 ... back to -8 again

Answer 11

i personally think the determinant method is simpler

Answer 12

Yeah but don't you learn that later....

Answer 13

phi, yours is not rref, effect but not rref :) just an echelon form

Answer 14

learn what later?

Answer 15

@amistre64 i personally think the determinant method is simpler

I think computationally its cost grows N! so for large N, you would want to do elimination.
but for this problem, definitely!

Answer 16

yes, just ref (not reduced row), sorry if that causes any confusion.

Answer 17

determinants

Answer 18

phi, right, i was blinded by the 3x3 of it ... lol

Answer 19

Ryan, depends onwhat level Remainder is at i spose

Answer 20

can u pls check this i 've didn't learn determinat method

Answer 21

@phi and @RyanL. CHECK THIS PLS!

Answer 22

why did you make the 3rd column all zeros?
if you are solving for
Mx=b
augment M with b
M | b
and row reduce

M in this case is just the 2 given vectors in the columns of M
so M is 3x2
M|b is 3x3

Answer 23

I i have this ,then wat must i do?