Matrix of $ T(ax^2+bx+c) = (3a + b)x^2 + (3a - 4b + c)x - a $ 
Let $T:P_3 \to P_3$ be defined by $$ T(ax^2+bx+c) = (3a + b)x^2 + (3a - 4b + c)x - a $$ Find the inverse of T.

I am assuming to find the inverse, I need to make a matrix of this transformation, then find the inverse of the matrix then turn it back into polynomial form.
Now to find the matrix for this transformation, I need a basis first and then put each part of the basis through the transformation right?
So I pick a basis of
$$ \mathfrak B = \{ 1,x,x^2 \} $$
However, I don't understand how each part of the base would go through the transformation. For instance,
$$ T(1) = -a $$
Or should it be this instead because I haven't given an $a$?
$$ T(1) = 0 $$
Edit:
Alright so the matrix of the linear transformation that I got is:
$$
\left[
\begin{matrix}
        0 & 0 & -1 \\
        1 & -4 & 3 \\
        0 & 1 & 3 \\
        \end{matrix}
\right]
$$
where I calculated the inverse and got:
$$
\left[
\begin{matrix}
        15 & 1 & 4 \\
        3 & 0 & 1 \\
        -1 & 0 & 0 \\
        \end{matrix}
\right]
$$
Is this correct so far? Now how would I convert this back into a polynomial?
 A: Try the simpler questions first:


*

*What are $T(1)$, $T(x)$ and $T(x^2)$?

*Do you know what is the matrix of $T$ with respect to your basis $\mathcal{B}$?


Then do the rest of the problem. 
A: I'll just respond to this part:

I am assuming to find the inverse, I need to make a matrix of this transformation

Not necessarily.  $T^{-1}$ is the function with the property
$$T^{-1}\left((3a + b)x^2 + (3a - 4b + c)x - a\right) = ax^2+bx+c$$
So define new variables $$a' = 3a+b, \\ b'=3a-4b+c, \\ c'=-a$$
Then from the third equation $a$ is clearly $-c'$, from the first equation $b$ is $a'+3c'$, and then plugging those both in the second equation, we get that $c$ is $b'+3c'+4(a'+3c') = 4a'+b'+15c'$.
So $$T^{-1}(a'x^2+b'x+c') = -c'x^2+(a'+3c')x+4a'+b'+15c'$$
Note: this does essentially come down to doing the exact same operations as you would do with a matrix.  And in more complicated cases you might have to construct a matrix to solve the system of linear equations, anyway.  But in principle you don't have construct $[T]_{\mathcal B}$.
A: Sure, you could use a basis method. However, why not try something a little simpler. First off, let's not consider $P_{3}$, let's work in $\mathbb{R}^3$ instead since it's an isomorphic space, and for the simple reason that it's something we're more used to, and convenient for matrix operations. Let's define the vector corresponding to a given polynomial $ax^2 + bx + c$ as $\begin{bmatrix}a\\b\\c\\\end{bmatrix}$. Then, let's take the transformation matrix as $$T = \begin{bmatrix}x_1 & x_2&x_3\\y_1 & y_2&y_3\\z_1 & z_2&z_3\end{bmatrix}$$ We know that $$T\begin{bmatrix}a\\b\\c\\\end{bmatrix} = \begin{bmatrix}3a+b\\3a-4b+c\\-a\\\end{bmatrix}$$ This means $$x_1a + x_2b+ x_3c = 3a+b$$ We know that certainly $x_1=3, x_2=1,$ and $c=0$ is a solution. Similarly, we find the values of $y_s$ and $z_s$. Then, our transformation matrix is$$T = \begin{bmatrix}3 & 1&0\\3 & -4&1\\-1 & 0&0\end{bmatrix}$$ We can easily find the inverse of this matrix using a gaussian method. This was done using software to save time, giving the matrix $$T^{-1} = \begin{bmatrix}0&0&-1\\1 & 0&3\\4 & 1&15\end{bmatrix}$$ You can then turn this into a polynomial transformation by the opposite of our initial method. $$\begin{bmatrix}0&0&-1\\1 & 0&3\\4 & 1&15\end{bmatrix}\begin{bmatrix}a\\b\\c\\\end{bmatrix}=\begin{bmatrix}-c\\a+3c\\4a+b+15c\\\end{bmatrix}$$ gives a polynomial transformation $T^{-1}(ax^2+bx+c)=-cx^2+(a+3c)x+4a+b+15$ Verify this against your original transformation.
