Linear Combinations with Polynomials Determine whether $q(x) = x^3 + x^2 - 3x + 2$ is a linear combination of $p_{1}(x) = x^3$,  $p_{2}(x) = x^2 + 3x$,  $p_{3}(x) = x^2 + 1$ in $P_{3}$ and, if so, find scalars $c_1, c_2, c_3$ such that $q(x) = c_{1} p_{1}(x) + c_{2} p_{2}(x) + c_{3} p_{3}(x)$. Are these scalars unique?
I know you're looking for the scalars in the linear combination but I'm not really sure how to start this. I'm not that familiar with using polynomials in this way. Thanks
 A: By inspection, you can see that $q=p_1-p_2+2p_3$, so the answer is yes.
To see that these scalars are unique, suppose not. Then there is some $a\ne 1$ or $b\ne -1$ or $c\ne 2$ and 
$$
ap_1+bp_2+cp_3=q=p_1-p_2+2p_3\implies ax^3+bx^2+3bx+cx^2+c=x^3+x^2-3x+2
$$ 
and equating terms of equal degree
$$
a=1\\
3b=-3\implies b=-1\\
c=2
$$
edit: Matrix solution, as requested. In terms of the standard basis $1,x,x^2,x^3$ of polynomials of degree at most 4, we have 
$$
p_1=\begin{bmatrix}0\\0\\0\\1\end{bmatrix},
p_2=\begin{bmatrix}0\\3\\1\\0\end{bmatrix},
p_3=\begin{bmatrix}1\\0\\1\\0\end{bmatrix}
$$
with the way I wrote $p_1$ being a hint of where this came from. 
Then you just need to solve 
$$
a\begin{bmatrix}0\\0\\0\\1\end{bmatrix}+b\begin{bmatrix}0\\3\\1\\0\end{bmatrix}+c\begin{bmatrix}1\\0\\1\\0\end{bmatrix}=\begin{bmatrix}2\\-3\\1\\1\end{bmatrix}
$$
which in matrix form means
$$
\begin{bmatrix}0&0&1\\0&3&0\\0&1&1\\1&0&0\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}2\\-3\\1\\1\end{bmatrix}
$$
can you take it from here, perhaps solving the linear system in augmented form and using what you about linear systems? Cough cough rank nullity. 
A: $q$ is a linear combination of $p_1, p_2, p_3$ iff these two matrices have the same rank:
$$
\pmatrix{
1 & 0 & 0 & 0 \\
0 & 1 & 3 & 0 \\
0 & 1 & 0 & 1 \\
}
\qquad
\pmatrix{
1 & 0 & 0 & 0 \\
0 & 1 & 3 & 0 \\
0 & 1 & 0 & 1 \\
1 & 1 & -3 & 2 \\
}
$$
This can be done by row reduction.
If you keep track of the row operations done on the second matrix, you'll find the coefficients, if they exist.
If the first matrix has full rank, then the coefficients are unique.
