Linear Algebra Problem row echelon 
Is there anyone could solve rhis question?
I try to minimize the matrix to row echelon form to find inconsistent and determinant of it but  i cannot reduce it some how 
try 1 = 
$$\left[\begin{array}{cccc|c} 2&-1 & -2a & 1 & b \\ -2 & a & -3 & 0 & 4\\ 2 & -1 & 2a+1 & a+1 & 0 \\ -2 &-1 & 1-2a & -2 & -2b-2 \end{array} \right]$$
$$\to r_1 +r_2$$
$$ \to -r_1+r_3$$
$$ \to r_1+r_4$$
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b \\ 0 & a-1 & -2a-3 & 1 & b+4\\ 0 & 0& 4a+1 & a & -b \\ 0 & 0 & 1-4a & -1 & -b-2 \end{array} \right]$$
$$\to r_3+r_4$$
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b\\ 0 & a-1 & -2a-3 & 1 & b+4\\ 0 & 0 & 4a+1 & a & -b\\ 0 & 0 & 2 & a-1 & -2b-2 \end{array}\right]$$
I am stuck?

Guide:
Great, now I will consider the following cases:
Case $1$: $a=1$. In this case, the matrix on the left is completely determined and it is certainly singular. 
$$\left[\begin{array}{cccc|c} 2 & -1 & -2 & 1 & b\\ 0 & 0 & -5 & 1 & b+4\\ 0 & 0 & 5 & 1 & -b\\ 0 & 0 & 2 & 0 & -2b-2 \end{array}\right]$$
Check if there are rows where the left hand side is zero but the right hand side is not. If such case exists, then there is no solution, otherwise, you have infinitely many solution. 
Case $2$:
Now, let's switch the third and fourth row:
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b\\ 0 & a-1 & -2a-3 & 1 & b+4 \\ 0 & 0 & 2 & a-1 & -2b-2 \\ 0 & 0 & 4a+1 & a & -b\end{array}\right]$$
Perform $-\frac{(4a+1)}2r_3+r_4$
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b\\ 0 & a-1 & -2a-3 & 1 & b+4 \\ 0 & 0 & 2 & a-1 & -2b-2 \\ 0 & 0 & 0 & -\frac{(4a+1)(a-1)}2+a & (4a+1)(b+1)-b\end{array}\right]$$
Find value of $a$ such that $(4,4)$ entry is zero, then depending on the value of $b$, you either have no solution of infinitely many solution. Otherwise, the solution is unique.
Edit: I realized there was a typo earlier, but fortunately it cancel out. when you first write down your matrix, it should be 
$$\left[\begin{array}{cccc|c} 2&-1 & -2a & 1 & b \\ -2 & a & -3 & 0 & 4\\ 2 & -1 & 2a+1 & a+1 & 0 \\ -2 & \color{red}1 & 1-2a & -2 & -2b-2 \end{array} \right]$$
