Find the values of c1 c2 and c3 so that c1 682 c2 4 4 0 c3

Find the values of c_1, c_2, and c_3 so that c_1 (6,-8,2) + c_2 (4, -4, 0) + c_3 (2, 0, 0) = (4, 8, 2).

Solution

c1*(6,-8,2) + c2*(4,-4,0)+c3(2,0,0) = (4,8,2)

c1*(6,-8,2) + c2*(4,-4,0)+c3(2,0,0) = (4,8,2)
(6c1+4c2+2c3 , -8c1-4c2 , 2c1) = (4,8,2)
equate both sides,
2c1=2
==> c1 = 1

-8c1-4c2 = 8
==> -8*1 - 4c2 = 8
==> -4c2 = 16
==> c2 = -4

6c1+4c2+2c3 = 4
==> 6*1 + 4*(-4) + 2*c3 = 4
==> 6-16 + 2*c3 = 4
==> c3 = 7

Answer:
c1 = 1
c2 = -4
c3 = 7