<s>[INST] A Gaussian elimination problem on Matrices Considering these linear equations :-
$-3x_1 - 3x_2 + 9x_3 = 12$
$2x_1 + 2x_2 - 4x_3 = -2$
$ -2x_2 - 4x_3 = -8$
Now to solve this problem I used the Gaussian elimination method.
I converted into an augmented matrix of the same
$$ \begin{bmatrix} -3 & -3 & 9 & 12 \\ 2 & 2 & -4 & -2 \\ 0 & -2 & -4 & -8 \end{bmatrix} $$
Then I performed these operations :-

*

*$[ \frac {-1} 3 R_1 \to R_1]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 2 & 2 & -4 & -2 \\ 0 & -2 & -4 & -8 \end{bmatrix} $$


*$[ R_2 - R_1 \to R_2]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 0 & 0 & -2 & 6 \\ 0 & -2 & -4 & -8 \end{bmatrix} $$


*$[ \frac {-1} {2} R_3 \to R_3]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 0 & 0 & -2 & 6 \\ 0 & 1 & 2 & 4 \end{bmatrix} $$


*$[ R_3 + R_2 \to R_2]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 0 & 1 & 0 & 10 \\ 0 & 1 & 2 & 4 \end{bmatrix} $$


*$[ R_1 - R_2 \to R_1]$ $[ R_3 - R_2 \to R_3]$
$$ \begin{bmatrix} 1 & 0 & -3 & -14 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 2 & -6 \end{bmatrix} $$


*$[ \frac {1} {2} R_3 \to R_3]$
$$ \begin{bmatrix} 1 & 0 & -3 & -14 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & -3 \end{bmatrix} $$


*And finally $[ R_1 + 3R_3 \to R_1]$
$$ \begin{bmatrix} 1 & 0 & 0 & -17 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & -3 \end{bmatrix} $$
$\therefore$ $x_1 = -17$, $x_2 = 10$ and $x_3 = -3$.
But the answer is given as $x_1 = 7$ , $x_2 = -2$ and $x_3 = 3$.
Where did I go wrong?
 [/INST] You went wrong in the second step (where you should also write $[R_{2}-2R_{1}\to R_{2}]$) and again in the last step.
With problems like this, you may want to consider leaving it be for a moment and coming back some time later. That makes it easier to spot your own mistakes.
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