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.
+128475 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 )
