Thread: Easy Gauss Jordan Elimination Matrix Problem

2x + 2y - z = 2
x - 3y + z = 0
3x + 4y - z = 1

Matrix form

2 2 -1 | 2
1 -3 1 | 0
3 4 -1 | 1

Could anyone sort of give me the steps to solve this? I'm having some trouble on it and I"m reviewing for an exam tomorrow. What sort of techniques should I be following? I normally just try to get a 1 a the top left, then a 0 in the 1st column 2nd row then a 0 in the 1st column 3rd row. But confused on what I should be looking for after that (I missed the lesson so I'm sort of teaching myself). All help appreciated!

well, let's DO that. note that this corresponds to "removing x from 2 of the equations (eliminating x in 2 equations)". subtracting 1/2 times row 1 from row 2 we get:

$\displaystyle \begin{bmatrix}2&2&-1&|&2\\0&-4&\frac{3}{2}&|&-1\\3&4&-1&|&1 \end{bmatrix}$

next, we subtract 3/2 row 1 from row 3 to get:

$\displaystyle \begin{bmatrix}2&2&-1&|&2\\0&-4&\frac{3}{2}&|&-1\\0&1&\frac{1}{2}&|&-2 \end{bmatrix}$

then we multiply row 1 by 1/2 to get:

$\displaystyle \begin{bmatrix}1&1&-\frac{1}{2}&|&1\\0&-4&\frac{3}{2}&|&-1\\0&1&\frac{1}{2}&|&-2 \end{bmatrix}$

rather then figure out some weird fractions, let's multiply row 2 by -1/4 now:

$\displaystyle \begin{bmatrix}1&1&-\frac{1}{2}&|&1\\0&1&-\frac{3}{8}&|&\frac{1}{4}\\0&1&\frac{1}{2}&|&-2 \end{bmatrix}$

this is apparently the point at which you get "stuck". but we just keep doing the same thing, except now our goal is to get rid of everything in column 2 except for the 1 in ROW 2. so our next step is to subtract row 2 from row 1:

$\displaystyle \begin{bmatrix}1&0&-\frac{1}{8}&|&\frac{3}{4}\\0&1&-\frac{3}{8}&|&\frac{1}{4}\\0&1&\frac{1}{2}&|&-2 \end{bmatrix}$

and next we subtract row 2 from row 3:

$\displaystyle \begin{bmatrix}1&0&-\frac{1}{8}&|&\frac{3}{4}\\0&1&-\frac{3}{8}&|&\frac{1}{4}\\0&0&\frac{7}{8}&|&-\frac{9}{4} \end{bmatrix}$

to get the 1 we want in row 3, column 3, multiply row 3 by 8/7:

$\displaystyle \begin{bmatrix}1&0&-\frac{1}{8}&|&\frac{3}{4}\\0&1&-\frac{3}{8}&|&\frac{1}{4}\\0&0&1&|&-\frac{18}{7} \end{bmatrix}$

now, it's going to get a bit messy, but we need to add 3/8 times row 3 to row 2:

$\displaystyle \begin{bmatrix}1&0&-\frac{1}{8}&|&\frac{3}{4}\\0&1&0&|&-\frac{5}{7}\\0&0&1&|&-\frac{18}{7} \end{bmatrix}$

and to finish off, add 1/8 times row 3 to row 1:

$\displaystyle \begin{bmatrix}1&0&0&|&\frac{3}{7}\\0&1&0&|&-\frac{5}{7}\\0&0&1&|&-\frac{18}{7} \end{bmatrix}$

so, if i have done my arithmetic correctly, x = 3/7, y = -5/7, z = -18/7 is a solution:

2(3/7) + 2(-5/7) - (-18/7) = 6/7 - 10/7 + 18/7 = 14/7 = 2 (check one)

3/7 - 3(-5/7) + (-18/7) = 3/7 + 15/7 - 18/7 = 0 (check two)

3(3/7) + 4(-5/7) - (-18/7) = 9/7 - 20/7 + 18/7 = 1 (check three, and verified as solution).

NOTE: i did the above row-reduction "idiot fashion" without looking for shortcuts. sometimes you can see easier ways to get the same thing done.

Thanks a lot! I greatly appreciate this!!