Invertible matrix($T:M_{2*2}(R) \rightarrow M_{2*2}(R)$ defined by T( $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$ 
$T:M_{2*2}(R) \rightarrow M_{2*2}(R)$ defined by T( $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$= $\bigl( \begin{smallmatrix} a+b & a \\ c & c+d \end{smallmatrix} \bigr)$. Justify whether T is invertible.

Once I get that it is one-to-one, can I apply the theorem in here to argue it is also onto?

Theorem: Let V and W be vector spaces of equal dimension, and let $T: V \rightarrow W$ be linear. Then the following are equivalent. a). T is one-to-one, b). T is onto, c). $rank(T)=dim(V)$

 A: As @user1551 mentioned in the comments, you may apply the stated theorem. Alternatively, one may observe that
$$
T\begin{pmatrix}
b & a-b\\
c & d-c
\end{pmatrix}
=
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix},
$$
to see that $T$ is onto.
A: This can also be seen directly from the definition of $T$, writing
$T \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = \begin{bmatrix} a + b & a \\ c & c + d \end{bmatrix} = \begin{bmatrix} x & y \\ z & w \end{bmatrix}, \tag 1$
comparing entries yields
$a + b = x, \tag 2$
$a = y, \tag 3$
$c = z, \tag 4$
$c + d = w; \tag 5$
thus,
$b = x - a = x - y, \tag 7$
$d = w - c = w - z; \tag 8$
therefore we may take
$T^{-1}\left (\begin{bmatrix} x & y \\ z & w \end{bmatrix} \right ) = \begin{bmatrix} y & x - y \\ z & w - z \end{bmatrix} = \begin{bmatrix} a  & b \\ c & d \end{bmatrix}, \tag 9$
which explicitly presents the inverse of $T$; indeed, we have from (9),
$T \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = T \left (\begin{bmatrix} y & x - y \\ z & w - z \end{bmatrix} \right )$
$= \begin{bmatrix} y + (x - y) & y \\ z & z + (w - z) \end{bmatrix} = \begin{bmatrix} x & y \\ z & w \end{bmatrix}. \tag{10}$
