Building a Linear Algebra Structure: Solving for Coefficients

[transgalactic]

the question is in the link:
http://img517.imageshack.us/img517/3830/70738563la6.gif

i know that's how i find coordinated(x,y,z) U=x*v1 +y*v2 +z+v3

but i don't know how to build this structure here?

 

[dirk_mec1]

In imageshack you see a link below which is named "hotlink for forums", copy THAT link here and the picture will be directly shown in your post.

 

[HallsofIvy]

You have a groupm, B, containing three members, ${b_1, b_2, b_3}$ and the vector space of all functions from B to R. (you say only "group of functions" but it must be a vector space for this to make sense.) You are also given a "basis" for that vector space defined by $g_i(b_j)$ equals 1 if i= j, 0 otherwise. You are asked to write g, defined by $g(b_1)= 1$, $g(b_2)= 4$, [itex]g(b_3)= 5.<br /> <br /> Okay, you must have $g(x)= a_1g_1(x)+ a_2g_2(x)+ a_3g_3(x)$. Set x= $b_1, b_2, b_3$ to get three very simple equations to solve for $a_1, a_2, a_3$.[/itex]

 

[transgalactic]

$g(b_1)= 1<br /> g(b_2)= 4<br /> g(b_3)= 5 <br /> g(x)= a_1g_1(x)+ a_2g_2(x)+ a_3g_3(x)$
$x=b1,b2,b3<br />$

$g(b1,b2,b3)= a_1g_1(b1,b2,b3)+ a_2g_2(b1,b2,b3)+ a_3g_3(b1,b2,b3)$

what is the next step
for constracting the equations

 

[HallsofIvy]

There is NO "$g(b2, b3,b3)$! Set x equal to b1, b2, and b3 separately to get three equations.

 

[transgalactic]

like this?

X(1,2,0)+y(0,1,2)+z(0,0,1)=(1,2,5)