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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}

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  • $\begingroup$ Welcome to Maths SX! There's an error: the last row show be $\;0\quad 0\quad b-11 \color{red}+4a\quad 2$. $\endgroup$ – Bernard Aug 11 '18 at 19:59
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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?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Daniel Fischer Aug 13 '18 at 17:45

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