Given the following linear transformation, find the matrix associated to $\varphi$ through a given base. Text of the exercise:
Consider the linear transformation $\varphi:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by: 
$\varphi(1,0,0)  = (0,1,0)\\ \varphi(0,1,1) = (1,1,0) \\\varphi(0,0,1) =(1,0,0)$ 
Is $B={(0,1,1),(0,1,0),(1,0,0)}$ a basis of $\mathbb{R}^3$? If the answer is yes, find the matrix $M_{\varphi}^B$ associated to $\varphi$ through $B$.
Reasoning:
$B$ is a basis of $\mathbb{R}^3$ since it is a set of three linearly independent vectors (proof is given by the fact that the matrix constituted by their coordinates is not singular) in a three-dimensional space.
I know from the definition of $\varphi$ the result of the transformation of two of the vectors of $B$; precisely $\varphi(1,0,0)=(0,1,0)$ and $\varphi(0,1,1)=(1,1,0)$. In order to infer the last one, I could use the definind property of a linear mapping, i.e. that a function $f:V\rightarrow W$ is linear if $\forall \textbf{v}_{1},\textbf{v}_{2}\in V$, $\forall\lambda_{1},\lambda_{2}\in \mathbb{K}$ then $f(\lambda_{1}\textbf{v}_{1}+\lambda_{2}\textbf{v}_{2}=\lambda_{1}f(\textbf{v}_{1})+\lambda_{1}f(\textbf{v}_{1})$, so that:
\begin{equation}
\varphi(0,1,0)=\varphi(0,1,1)-\varphi(0,0,1)=(1,1,0)-(1,0,0)=(0,1,0)
\end{equation}
The asked matrix is thus:
\begin{equation}
M_{\varphi}^B=\begin{bmatrix}
1&0&0\\1&1&1\\0&0&0
\end{bmatrix}
\end{equation}
I'd greatly appreciate a feedback. Thank you all!

Indeed B is a basis for exactly the reason you say.
$e_1 = (0,1,1), e_2 = (0,1,0), e_3 = (1,0,0)$
$\varphi e_1 = \varphi (0,1,1) = (1,1,0) = e_2 + e_3\\
\varphi e_2 =  \varphi (0,1,1)- \varphi (0,0,1)= (1,1,0) - (1,0,0) = (0,1,0) = e_2\\
\varphi e_3 = \varphi (1,0,0)= (0,0,1) = e_1 - e_2$
$M = \begin{bmatrix} 0&0&1\\1&1&-1\\1&0&0\end{bmatrix}$
