Ytukyg

Determine which of the following vectors p(t) is a linear combination of:$$p_1=t^2-t$$ $$p_2=t^{2}-2t+1$$ $$p_3=-t^{2}+1$$

a) $$p(t)=3t^{2}-3+1$$
b) $$p(t)=t^{2}-t+1$$
c) $$p(t)=t+1$$
d) $$p(t)=2t^{2}-t-1$$

Ok. This is why I'm doing... For example to check the first one, the letter a, i'm using the scalars $$x,y,z$$.

$$p(t)=xp_1+yp_2+zp_3$$

$$3t^2-3t+1=x(t^2-t)+y(t^2-2t+1)+z(-t^2+1)$$

$$3t^2-3t+1=xt^2-xt+yt^2-2ty+y-t^2z+z$$

$$3t^2-3t+1=xt^2+yt^2-t^2z-xt-2ty+y+z$$

$$3t^2-3t+1=(x+y-z)t^2-(x+2y)t+(y+z)$$

Then I get and this equation system:

$$x+y-z=3$$
$$x+2y=3$$
$$y+z=1$$

which doesn't have a solution and then I conclude that $$p(t)$$ is not a linear combination of the vectors $$p_1,p_2,p_3$$.

It is my analysis right? I did the same with the rest and when having the equation system non of them have solution so I conclude that non of them are linear combination. But I want to be sure that what I'm doing is right.

I'm teaching myself linear algebra and any help from the math community will be appreciated.

We have

$$p_1(1)=p_2(1)=p_3(1)=0$$
thus you will have
$$p(1)=alpha p_1(1)+beta p_2(1)+gamma p_3(1)=0$$

the only one satisfying this condition is the last :(d).
$$p(t)=2t^2-t-1=p_1(t)-p_3(t)$$