Answer
False
Work Step by Step
Assume $p=a_0+a_1x+a_2x^2+a_3x^3 \in P_3(R)$
We have a linear transformation $P_3(R)\rightarrow M_{32}(R)$
obtain: $T(p)=T(a_0+a_1x+a_2x^2+a_3x^3)=\begin{bmatrix}
a_0 & a_1\\
a_2 & a_3\\
0 & 0
\end{bmatrix}$
then $Ker(T)=\{p \in P_3(R):T(p)=0\}\\
=\{a_0+a_1x+a_2x^2+a_3x^3:\begin{bmatrix}
a_0 & a_1\\
a_2 & a_3\\
0 & 0
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0 & 0\\
0 & 0
\end{bmatrix}\}\\
\rightarrow a_0=a_1=a_2=a_3=0\\
\rightarrow Ker(T)=\{0\}$
Hence, there is one-to-one linear transformation $T$
