# Math Help - I need help with a Guass Jordan elimination problem

1. ## I need help with a Guass Jordan elimination problem

3 8 -1 :-18
2 1 5 : 8
2 4 2 :-4

2. Originally Posted by keisanhen
3 8 -1 :-18
2 1 5 : 8
2 4 2 :-4
$A=\left[\begin{array}{ccc|c} 3 & 8 & -1 & -18 \\ 2 & 1 & 5 & 8 \\ 2 & 4 & 2 & -4 \end{array}\right]$

$A_{\begin{array}{l} r_1\leftrightarrow\frac{1}{2}r_3 \end{array}}=\left[\begin{array}{ccc|c} 1 & 2 & 1 & -2 \\ 2 & 1 & 5 & 8 \\ 3 & 8 & -1 & -18 \end{array}\right]$

$A_{\begin{array}{l} r_1\leftrightarrow\frac{1}{2}r_3 \\ r_3\rightarrow r_3-3r_1 \end{array}}=\left[\begin{array}{ccc|c} 1 & 2 & 1 & -2 \\ 2 & 1 & 5 & 8 \\ 0 & 2 & -4 & -12 \end{array}\right]$

$A_{\begin{array}{l} r_1\leftrightarrow\frac{1}{2}r_3 \\ r_3\rightarrow r_3-3r_1 \\ r_2\rightarrow\frac{2}{3}r_1-\frac{1}{3}r_2 \end{array}}=\left[\begin{array}{ccc|c} 1 & 2 & 1 & -2 \\ 0 & 1 & -1 & -4 \\ 0 & 2 & -4 & -12 \end{array}\right]$

$A_{\begin{array}{l} r_1\leftrightarrow\frac{1}{2}r_3 \\ r_3\rightarrow r_3-3r_1 \\ r_2\rightarrow\frac{2}{3}r_1-\frac{1}{3}r_2 \\ r_3\rightarrow r_2-\frac{1}{2}r_3 \end{array}}=\left[\begin{array}{ccc|c} 1 & 2 & 1 & -2 \\ 0 & 1 & -1 & -4 \\ 0 & 0 & 1 & 2 \end{array}\right]=B$

$B=\left[\begin{array}{ccc|c} 1 & 2 & 1 & -2 \\ 0 & 1 & -1 & -4 \\ 0 & 0 & 1 & 2 \end{array}\right]$

$B_{\begin{array}{l}r_1\rightarrow r_1-r_3 \end{array}}=\left[\begin{array}{ccc|c} 1 & 2 & 0 & -4 \\ 0 & 1 & -1 & -4 \\ 0 & 0 & 1 & 2 \end{array}\right]$

$B_{\begin{array}{l}r_1\rightarrow r_1-r_3 \\ r_2\rightarrow r_2+r_3 \end{array}}=\left[\begin{array}{ccc|c} 1 & 2 & 0 & -4 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 2 \end{array}\right]$

$B_{\begin{array}{l}r_1\rightarrow r_1-r_3 \\ r_2\rightarrow r_2+r_3 \\ r_1\rightarrow r_1-2r_2 \end{array}}=\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 2 \end{array}\right]$

3. Thanks very very much

4. Originally Posted by keisanhen
What made you choose the fraction 2/3 and 1/3 for line two, which was 2 1 5 8, to solve
Well, to get the matrix into row echelon form, we want to get zeros in the bottom-left triangle. The only way to do that with elementary row operations is to combine each row with some proportion of another row. When tackling the first-column entry of row 2, I found that row 3 wouldn't work, since its first-column value was zero. The only remaining option was row 1. $r_2\rightarrow2r_1-r_2$ is our first choice, but we can kill two birds with one stone by dividing all that by 3:

$r_2\rightarrow\frac{1}{3}(2r_1-r_2)$

...or...

$r_2\rightarrow\frac{2}{3}r_1-\frac{1}{3}r_2$

This not only makes the first-column entry a zero, but it makes the second-column entry a 1, which is also necessary for row echelon form.