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!