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No matter what I do I can't seem to get this in the proper form. Here is the system:

$$ \left\{ \begin{aligned} 3x+3y+12z&=6 \\ 3x_1+x_2-2x_3&=2 \\ 2x_1+2x_2+x_3&=10 \\ -x+2y+8z&=4 \end{aligned} \right. $$

Here is my base matrix:

$$ \left[ \begin{array}{cccc|c} 3&3&12&6\\ 1&1&4&2 \\ 2&5&20&10 \\ -1&2&8&4 \end{array} \right]$$

$\left(\frac{1}{3}\right)R_1->R_1$

$$ \left[ \begin{array}{cccc|c} 1&1&4&2\\ 1&1&4&2 \\ 2&5&20&10 \\ -1&2&8&4 \end{array} \right]$$

$(-2)R_1+R_3->R_3$ $$ \left[ \begin{array}{cccc|c} 1&1&4&2\\ 1&1&4&2 \\ 0&3&12&6 \\ -1&2&8&4 \end{array} \right]$$

$R_2<->R_4$ $$ \left[ \begin{array}{cccc|c} 1&1&4&2\\ -1&2&8&4 \\ 0&3&12&6 \\ 1&1&4&2 \end{array} \right]$$

$R_2<->R_3$ $$ \left[ \begin{array}{cccc|c} 1&1&4&2\\ 0&3&12&6 \\ -1&2&8&4 \\ 1&1&4&2 \end{array} \right]$$

$R_3+R_4->R_4$ $$ \left[ \begin{array}{cccc|c} 1&1&4&2\\ 0&3&12&6 \\ -1&2&8&4 \\ 0&1&4&2 \end{array} \right]$$

$R_2<->R_3$, $R_1+R_2->R_2$ $$ \left[ \begin{array}{cccc|c} 1&1&4&2\\ 0&-1&-4&-2 \\ 0&3&12&7 \\ 0&1&4&2 \end{array} \right]$$

$\left(\frac{1}{3}\right)R_3->R_3$ $$ \left[ \begin{array}{cccc|c} 1&1&4&2\\ 0&-1&-4&-2 \\ 0&1&4&2 \\ 0&1&4&2 \end{array} \right]$$

$R_1<->R_2$ $$ \left[ \begin{array}{cccc|c} 0&-1&-4&-2\\ 1&1&4&2 \\ 0&1&4&2 \\ 0&1&4&2 \end{array} \right]$$

$R_1+R_2->R_2$ $$ \left[ \begin{array}{cccc|c} 0&-1&-4&-2\\ 1&0&0&0 \\ 0&1&4&2 \\ 0&1&4&2 \end{array} \right]$$

$R_1<->R_2$ $$ \left[ \begin{array}{cccc|c} 1&0&0&0 \\ 0&-1&-4&-2 \\ 0&1&4&2 \\ 0&1&4&2 \end{array} \right]$$

$R_2<->R_3$ $$ \left[ \begin{array}{cccc|c} 1&0&0&0\\ 0&1&4&2 \\ 0&-1&-4&-2 \\ 0&1&4&2 \end{array} \right]$$

$R_2+R_3->R_3$ $$ \left[ \begin{array}{cccc|c} 1&0&0&0\\ 0&1&4&2 \\ 0&0&0&0 \\ 0&1&4&2 \end{array} \right]$$

I stopped here because I felt like I was going to be going into a circle and I felt like I've done way too many steps. Please help!

Sorry for any mistakes, it took me awhile to type this up.

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    $\begingroup$ The final result is correct! Eliminate the last row (you have two rows of zero). Great job! $\endgroup$ – Amzoti Jan 25 '14 at 19:04
  • $\begingroup$ @Amzoti wow, thanks. $\endgroup$ – hax0r_n_code Jan 25 '14 at 19:12

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