Gaussian elimination in linear equations

Question

So I have problems with solving this question, can someone please help me out with this one? I really don't need a direct solution I just need to know how should I do it, or if u are generous enough can you explain it step by step?

Determine with the Gaussian elimination the solution amounts of the following systems of linear equations - over $\mathbb R$

i) $−x_1 + 7x_2 − x_3 = 5$ ∧ $4x_1 − x_2 + x_3 = 1$ ∧ $5x_1 − 3x_2 + x_3 = −1$;

Answer 1

$$(1)\ -x_1+7x_2-x_3=5$$

$$(2)\ 4x_1-x_2+x_3=1$$

$$(3)\ 5x_1-3x_2+x_3=-1$$

$$(1)+(2)\ \ 3x_1+6x_2=6$$

$$(1)+(3)\ \ 4x_1+4x_2=4$$

This implies

$$x_1+2x_2=2$$

$$x_1+x_2=1$$

Subtracting gives $x_2=1$. It follows $x_1=0$ and $x_3=2$ So, the solution is $(0/1/2)$

This is the easiest way to solve the equation.

The Gauss-algorithm works as follows

-1 7 -1 | 5

4 -1 1 | 1

5 -3 1 | -1

New second row = 4 times first row + old second row New third row = 5 times first row + old third row

-1 7 -1 | 5

0 27 -3 | 21

0 32 -4 | 24

New third row : 32 times second row - 27 times third row

-1 7 -1 | 5

0 27 -3 | 21

0 0 12 | 24

We have reached triangle shape and we see $12x_3=24$ implying $x_3=2$ Inserting $x_3$ in $27x_2-3x_3=21$, we get $27x_2-6=21$, implying $x_2=1$. Finally, we get $x_1=0$ by inserting $x_2$ and $x_3$ in $-x_1+7x_2-x_3=5$ implying $-x_1+5=5$.

We have the unique solution $(0/1/2)$ again. The number of sulutions is $1$.