look at the figure below ..
![enter image description here][1]



I'm trying to follow this step-by-step but I couldn't know how this is evaluated?

I tried by matrix representation to understand the concept but I couldn't .. as following:

$$  
\begin{pmatrix}
ac & ad \\
bc & bd 
\end{pmatrix}
$$
Then After apply first Controlled-NOT we get 
$$
\begin{pmatrix}
ac & ad \\
bd & bc 
\end{pmatrix}
$$
Then After apply Second Controlled-NOT we get
$$
\begin{pmatrix}
ac & bc \\
bd & ad 
\end{pmatrix}
$$
Then After apply Third Controlled-NOT we get
$$
\begin{pmatrix}
ac & bc \\
ad & bd 
\end{pmatrix}
$$

I tried to follow how this was managed but I couldn't. Because I didn't find any thing that I can multiply to get the previous matrices.. (in some point I thought Pauli matrix X gate would manage that but it doesn't)

I tried to use the Controlled-NOT 4 by 4 matrix as shown in below 


so I tried to multiply this matrix to 4-by-1 matrix such as 
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0&  1\\
0 & 0 & 1&  0
\end{pmatrix}
\cdot
\begin{pmatrix}ac \\ac \\bc \\bd \end{pmatrix} =
\begin{pmatrix}ac \\ad \\bd \\bc \end{pmatrix}
$$

Now, taking the result and multiply bu Controlled-NOT again, we get
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0&  1\\
0 & 0 & 1&  0
\end{pmatrix}
\cdot
\begin{pmatrix}ac \\ad \\bd \\bc \end{pmatrix} =
\begin{pmatrix}ac \\ad \\bc \\bd \end{pmatrix}
$$
it seems that I did nothing, since the result is the same as the initial one ..
so, can you convince me by algebraic (matrix method) that by using Controlled-NOT, then I can get the result as the shown in the figure above ..

Thank you ..
Reference: the figure is taken from this [paper][2]
4-by-4-matrix is called Controlled-NOT gate see this [page][3]


  [1]: https://i.sstatic.net/TbKmT.png
  [2]: http://www-inst.eecs.berkeley.edu/~cs191/fa07/lectures/lecture9_fa07.pdf
  [3]: http://www.quantiki.org/wiki/CNOT