# Does this guassian elimination have a solution?

I was asked to find the following solutions using guassian elimination, but I was unsure of my answers since it became quite messy but the variables still somehow fit:

$$\left[\begin{array}{ccc|c} 3&4&1&7\\ 0&1&3&7\\ 3&3&-1&9\\ 1&2&3&8 \end{array}\right]$$

\begin{align*} &r_1\leftarrow r_1\\ &r_2\leftarrow r_2\\ &r_3\leftarrow r_3-1.5r_1\\ &r_4\leftarrow r_4 \end{align*}

Using the above process:

$$\left[\begin{array}{ccc|c} 2&4&1&7\\ 0&1&3&7\\ 0&-3&-2.5&-1.5\\ 1&2&-3&8 \end{array}\right]$$

\begin{align*} &r_1\leftarrow r_1\\ &r_2\leftarrow r_2\\ &r_3\leftarrow r_3-3r_4\\ &r_4\leftarrow r_4-0.5r_1 \end{align*}

$$\left[\begin{array}{ccc|c} 2&4&1&7\\ 0&1&3&7\\ 0&-3&-2.5&-1.5\\ 0&0&2.5&4.5 \end{array}\right]$$

\begin{align*} &r_1\leftarrow r_1\\ &r_2\leftarrow r_2\\ &r_3\leftarrow r_3+3r_2\\ &r_4\leftarrow r_4-\frac{r_3}{2.6}&\text{unsure about this step} \end{align*}

$$\left[\begin{array}{ccc|c} 2&4&1&7\\ 0&1&3&7\\ 0&0&6.5&19.5\\ 0&0&0&-3 \end{array}\right]$$

• @Dan: The original, accessed by clicking on edit, was actually fairly readable. What people don’t realize is that long breaks disappear. – Brian M. Scott Oct 25 '14 at 4:21