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Consider the 1 by 3 vectors 

v = <1 , 2 , 1>  w = <1 , 4 , 5>  x = <-1 , 6 , 15>.

 

 If they aren't, find coefficients a, b, and c, not all 0, such that

 

a <1 , 2 , 1>  + b<1 , 4 , 5> + c<-1 , 6 , 15> = <0 , 0 , 0>.

 

unless they are linearly independent.

 Jul 15, 2019
 #1
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We can see if we can find a solution for this system

 

a   +  b   - c    =  0       (1)   

2a  + 4b + 6c  = 0       (2)

a    + 5b  + 15c = 0     (3)

 

Multiply  (1) by 6   and add to (2)  and we get that

 

8a + 10b  = 0    (4)

 

Multiply (1) by 15 and add to (3)

16a + 20b = 0    (5)

 

(5)  is a multiple of (4).....so these are dependent

  

8a  = -10b

a = -10/8  b

a = -5/4 b

b = -4/5a

 

And 

a + b - c  = 0

a - (4/5)a  - c  = 0

(1/5)a  =  c

 

So...if we let a  =  5    then  b = (-4/5)(5) = -4   and c = (1/5)(5)  = 1

 

Check

 

5 ( 1, 2, 1)   - 4 ( 1, 4 , 5 )  + 1 ( - 1, 6 , 15 )   =  ( 0 , 0 , 0 ) 

 

 

 

cool cool cool

 Jul 15, 2019

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