(a) We will compute the kernel of $T$. Suppose that $a+bx+cx^2\in\krn{T}$. Then
\begin{equation*}
\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
=\lt{T}{a+bx+cx^2}
=
\begin{bmatrix}
a+2b-2c & 2a+2b \\
-a+b-4c & 3a+2b+2c
\end{bmatrix}
\end{equation*}
and matrix equality (Theorem ME) yields the homogeneous system of four equations in three variables,
\begin{align*}
a+2b-2c&=0\\
2a+2b&=0\\
-a+b-4c&=0\\
3a+2b+2c&=0\\
\end{align*}
The coefficient matrix of this system row-reduces as
\begin{equation*}
\begin{bmatrix}
1 & 2 & -2 \\
2 & 2 & 0 \\
-1 & 1 & -4 \\
3 & 2 & 2
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 2 \\
0 & \leading{1} & -2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
From the existence of nontrivial solutions to this system, we can infer nonzero polynomials in $\krn{T}$. By Theorem KILT we then know that $T$ is not injective.
(b) Since $3=\dimension{P_2}<\dimension{M_{22}}=4$, by Theorem SLTD $T$ is not surjective.
(c) Since $T$ is not surjective, it is not invertible by Theorem ILTIS.