For the following V is a threedimensional space of tradition

For the following, V is a three-dimensional space of traditional vectors with standard basis S={i, j,k) . (If you prefer, use ei, e2,e3] or {x, y,z] .) Also, let B= {b1 , b2 , b3 } where

Solution

let x i + y j + z k = a b1 + b b2 + c b3

in

(x,y,z) = a(1,2,0) + b(0,3,-1) + c(2,-3,0)

x = a + 2c

y = 2a +3b-3c

z = -b

solving we get

b = -z

a + 2c = x

2a -3c = y+3z

so

2a + 4c = 2x

2a -3c = y+3z

7c = (2x-y-3z)

or c = 1/7 (2x-y-3z)

a = x -2c = x - 2*1/7 (2x-y-3z)

= 1/7 (3x+2y+6z)

(x,y,z)|B = {1/7 (3x+2y+6z), -z,1/7 (2x-y-3z)}

now

a) i|S = (1,0,0) , similarly j |S - (0,1,0) and k|S - (0,0,1)

b1|S = (1,2,0)

b2|S - (0,3,-1)

b3|S = (2,-3,0)

i|B ,

(x,y,z)|B = {1/7 (3x+2y+6z), -z,1/7 (2x-y-3z)}

here x = 1 , y = 0 , z = 0

= (3/7,0,2/7)

similarly others can be calculated by puting the values of x,y,z

so