A Gaussian elimination problem on Matrices

Question

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 :-

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

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

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

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

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

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

7. 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?

Answer 1

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.