I am stuck with this linear system:
\begin{cases} x_1 + x_3 = 0 \\ ax_1 + x_2 + 2x_3 = 0 \\ 3x_1 + 4x_2 + bx_3 = 2 \end{cases}
My augmented matrix so far looks like this: R2- aR1 R3- 3R1 and then R3-4R2
\begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 2-a & 0 \\ 0 & 0 & b-3-8-4a & 2 \end{bmatrix}
Hint: From the first equation we get $$x_3=-x_1$$ so we can write
$$ax_1+x_2-2x_1=0$$ and
$$3x_1+4x_2-bx_1=2$$
Now we will eliminate $$x_2=2x_1-ax_1$$ so we will get
$$3x_1+4(2x_1-ax_1)-bx_1=2$$ or
$$x_1(11-4a-b)=2$$
Can you finish?