Math Help - Gaussian Elimination - 3 Simultaneous equations

1. Gaussian Elimination - 3 Simultaneous equations

Hi all,

Really, really struggling with this question.
I've tried using a few examples but I cant seem to get narrow down 3 terms to just 1.
I always end up with 2 letters/terms remaining.

The 3 equations I have are:

7i + 8ii + 7iii = 2 (Equation 1)
6i + 7ii + 5iii = 8 (Equation 2)
6i + 9ii + 6iii = 2 (Equation 3)

So I need to work out the values for i, ii and iii.

I tried using the 3 step procedure as described in a text book I have. But it doesn't seem to work.

I hope someone can help.

2. Re: Gaussian Elimination - 3 Simultaneous equations

instead of i, ii and iii i will use a,b, and c. so our 3 equations become:

7a + 8b + 7c = 2
6a + 7b + 5c = 8
6a + 9b + 6c = 2.

subtract equation 2 from equation 3 to obtain:

2b + c = -6 (*)

now, we need another equation with just b and c in it, to see if we can eliminate another variable. since we want the "a" terms to cancel, multiply equation 1 by 6, and equation 2 by 7:

42a + 48b + 42c = 12 (equation 1a)
42a + 49b + 35c = 56 (equation 2a)

then subtract equation 1a from equation 2a to get:

b - 7c = 44 (**), and now we multiply (**) by 2 to get:

2b - 14c = 88. subtract (*) from this, and we have:

-15c = 94, so c = -94/15.

using (*) we have:

2b - 94/15 = -6
b - 47/15 = -3
b = -3 + 47/15 = -45/15 + 47/15 = 2/15

and finally, using equation 1, we have:

7a + 8(2/15) + 7(-94/15) = 2, so:

a + 16/105 - 94/15 = 2/7 thus:

a = 2/7 - 16/105 + 94/15 = 30/105 - 16/105 + 658/105 = 672/105 = 96/15 = 32/5.

now let's see if our solution is correct (the first two times i did this, i made errors, so don't feel bad if you did, too):

7(32/5) + 8(2/15) + 7(-94/15) = 224/5 + 16/15 - 658/15 = 4704/105 + 112/105 - 4606/105 = 210/105 = 2 (so equation 1 checks out).

6(32/5) + 7(2/15) + 5(-94/15) = 192/5 + 14/15 - 94/3 = 576/15 + 14/15 - 470/15 = 120/15 = 8 (so equation 2 checks out).

6(32/5) + 9(2/15) + 6(-94/15) = 192/5 + 6/5 - 188/5 = 10/5 = 2 (all three equations check out).

so a = 32/5, b = 2/15, c = -94/15 is indeed the correct solution (with some rather ugly arithmetic attached).

3. Re: Gaussian Elimination - 3 Simultaneous equations

Thanks Deveno!

That is perfect!!

Going back over my original attempts I realise that NOT using fractions was a mistake! The decimal numbers got VERY messy.

Thanks again, for taking the time to write it all out. I actually understand the process better now.

Chris

4. Re: Gaussian Elimination - 3 Simultaneous equations

Hello, chrisa112!

Didn't you say Gaussian elimination?

$\begin{array}{ccc|c}7a + 8b + 7c &=& 2 \\ 6a + 7b + 5c &=& 8 \\ 6a + 9b + 6c &=& 2 \end{array}$

We have: . $\left|\begin{array}{ccc|c}7&8&7&2 \\ 6&7&5&8 \\ 6&9&6&2 \end{array}\right|$

$\begin{array}{c}R_1-R_2 \\ \\ R_3-R_2 \end{array}\left|\begin{array}{ccc|c}1&1&2&\text{-}6 \\ 6&7&5&8 \\ 0&2&1&\text{-}6 \end{array}\right|$

$\begin{array}{c} \\ R_2-6R_1 \\ \\ \end{array}\left|\begin{array}{ccc|c}1&1&2&\text{-}6 \\ 0&1&\text{-}7&44 \\ 0&2&1&\text{-}6 \end{array}\right|$

$\begin{array}{c}R_1-R_2\ \\ R_3-2R_2 \end{array}\left|\begin{array}{ccc|c}1&0&9&\text{-}50 \\ 0&1&\text{-}7&44 \\ 0&0&15&\text{-}94 \end{array}\right|$

. . $\begin{array}{c}\\ \\ \frac{1}{15}R_3 \end{array}\left|\begin{array}{ccc|c}1&0&9&\text{-}50 \\ 0&1&\text{-}7&44 \\ 0&0&1&\text{-}\frac{94}{15}\end{array}\right|$

$\begin{array}{c}R_1 - 9R_3 \\ R_2 + 7R_3 \\ \\ \end{array}\left|\begin{array}{ccc|c}1&0&0&\frac{3 2}{5} \\ 0&1&0&\frac{2}{15} \\ 0&0&1&\text{-}\frac{94}{15} \end{array}\right|$

5. Re: Gaussian Elimination - 3 Simultaneous equations

What Deveno did is also "Gaussian Elimination" as well as your row-reduction.

6. Re: Gaussian Elimination - 3 Simultaneous equations

Yes Deveno had solved it very smartly and this is a “Gaussian Elimination” even I got agreed with Hallsoflvy. Good job done by Deveno.