solve the system using the Gaussian method

Question

$$ x_1-x_2+3x_3+4x_4+x_5=1 $$ $$ 2x_1+7x_2+5x_3-2x_4 =4 $$ $$ x_1+8x_2+2x_3-6x_4-x_5=2 $$

Answer 1

I suggest you follow your work but without using a matrix to try to make you understand what was said to you in the comments. You should make a minimum of sentences to better understand what you are doing and where are the possible errors that are shown to you. I suggest for example:

Suppose $\color{blue}{\text{there are five real numbers}}$ which we will denote by $x_1, x_2,x_3,x_4$ and $x_5$ such that $$\begin{cases}x_1-x_2+3x_3+4x_4+x_5=1 \color{red}{ (I)}\\2x_1+7x_2+5x_3-2x_4+0x_5=1\color{red}{ (II)}\\x_1+8x_2+2x_3-6x_4-x_5=2\color{red}{ (III)}\end{cases}$$ So like you did, $$\begin{cases}x_1-x_2+3x_3+4x_4+x_5=1 \color{red}{ (I)}\color{green}{(I)}\\0x_1+9x_2-x_3-10x_4-2x_5=-1\color{red}{ (II)-2\times(I)}\color{green}{(II)}\\0x_1+9x_2-x_3-10x_4-2x_5=1\color{red}{ (III)-(I)\color{green}{(III)}}\end{cases}$$ Then, as @Paul told you, according to $\color{green}{(II)}$ and $\color{green}{(III)}$, $-1=9x_2-x_3-10x_4-2x_5=1.$ Then $-1=1$, what is wrong. So our $\color{blue}{\text{starting assumption}}$ is not possible. In other words, there is no $(x_1,x_2,x_3,x_4,x_5)\in\mathbb R^5$ such that $\begin{cases}x_1-x_2+3x_3+4x_4+x_5=1 \\2x_1+7x_2+5x_3-2x_4+0x_5=1\\x_1+8x_2+2x_3-6x_4-x_5=2\end{cases}$

Otherwise written, the set $S$ defined by $S:=\left\{(x_1,x_2,x_3,x_4,x_5)\in\mathbb R^5:\begin{cases}x_1-x_2+3x_3+4x_4+x_5=1 \\2x_1+7x_2+5x_3-2x_4+0x_5=1\\x_1+8x_2+2x_3-6x_4-x_5=2\end{cases}\right\}=\varnothing$