Solve different solutions for linear system I am currently in a linear algebra class and I have to answer the following questions, for what values of $a$ and $b$ does the system below have:
a) No solution 
b) Only one solution 
c) Infinitely many solutions 
\begin{cases} 
    x_{2} + 3x_{3} = 1 \\
    x_{1} + 2x_{2} + 6x_{3} = 1 \\
    x_{2} - 6x_{3} = -1 \\
    2x_{1} + 2x_{2} + ax_{3} = b
\end{cases}
To attempt this I tried to reduce the augmented matrix to echelon form and ended up with this
\begin{bmatrix} 
 1 & 2 & 6 & 1 \\ 
 0 & 2 & 3 & 1 \\ 
 0 & 0 & -9 & -2 \\
 0 & 0 & a & b + 1
\end{bmatrix}
From my understanding, for this particular system to have infinitely many solutions, both the third and the forth row has to be only zeros. Because since $a$ is in the third column, this is the only way to achieve $x_{3}$ being a free variable. The problem is that I don't know how to reduce this system in such a way and wondered if someone could help! I think if I understand how to arrange the matrix I will be able to answer the other questions. Thank you!
 A: you have to write augmented matrix. 
then you can to perform row operations to convert this into   echelon form
this link can help .
https://www.wikihow.com/Reduce-a-Matrix-to-Row-Echelon-Form
A: The given system has augmented matrix $$\begin{bmatrix}0&1&3&|&1\\1&2&6&|&1\\0&1&-6&|&-1\\2&2&a&|&b\end{bmatrix}\xrightarrow{(1)}\begin{bmatrix}0&1&3&|&1\\1&2&6&|&1\\0&1&-6&|&-1\\0&-2&a-12&|&b-2\end{bmatrix}\xrightarrow{(2)}\begin{bmatrix}0&1&3&|&1\\1&0&0&|&-1\\0&0&-9&|&-2\\0&0&a-6&|&b\end{bmatrix}$$
where the steps are $$\begin{align}&(1)\qquad -2R_2+R_4\mapsto R_4\\&(2)\qquad \begin{cases}-2R_1+R_2\mapsto R_2\\-R_1+R_3\mapsto R_3\\2R_1+R_4\mapsto R_4\end{cases}\end{align}$$
The system is inconsistent (no solutions) when there is a nonzero row in the coefficient matrix with a zero entry in the augmented vector on the right hand side. The system has one solution when the system is consistent and the matrix has rank $3$. Otherwise, there are infinitely many solutions.
One can perform a final step via $$(3)\qquad\begin{cases}-3R_3+R_1\mapsto R_1\\(6-a)R_3+R_4\mapsto R_4\end{cases}$$
$$\begin{bmatrix}0&1&3&|&1\\1&0&0&|&-1\\0&0&1&|&\frac{2}{9}\\0&0&a-6&|&b\end{bmatrix}\xrightarrow{(3)}\begin{bmatrix}0&1&0&|&\frac{1}{3}\\1&0&0&|&-1\\0&0&1&|&\frac{2}{9}\\0&0&0&|&b+\frac{2}{9}(6-a)\end{bmatrix}$$
A: Let's start from here:
\begin{bmatrix} 
 1 & 2 & 6 & 1 \\ 
 0 & 2 & 3 & 1 \\ 
 0 & 0 & -9 & -2 \\
 0 & 0 & a & b + 1
\end{bmatrix}
For equations 1 to 3 we calculate determinant:
$$
\begin{vmatrix} 
 1 & 2 & 6 \\ 
 0 & 2 & 3 \\ 
 0 & 0 & -9  \\
\end{vmatrix}=18>0
$$
This means the first $3$ equations have a unique solution.
Our system is overdetermined, so if the fourth equation is identical to the third one, the whole system has 1 unique solution and otherwise it has no solution.

If $a=-9$ AND $b+1=-2$ then we have 1 solution. Otherwise, we have no solution.

A: There's a general result for this:

Let $A\,\mathbf  x =\mathbf b$ be a  (possibly non-homogeneous) linear system of size $m\times n$ over the field $K$. We'll denote  $\;[A\mid \mathbf b]$ the augmented matrix. This linear system   has
  
  
*
  
*no solution  if $\;\DeclareMathOperator{\rank}{rank} \rank A<\rank A\mid\mathbf b$;
  
*solutions if $\;\rank A=\rank A\mid\mathbf b$.
Furthermore, if there are solutions, it is a single solution if $A$ has maximal rank, i.e. if $\rank A=\min(m,n)$. It has an infinity of solutions if $\rank A<\min(m,n)$.
In  all cases, the set of solutions is an affine subspace of $K^n$ with dimension $\operatorname{codim} A$.

