082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

8.10 Gaussian elimination 289
anything we please, thatis
y = λ
λisour free choice
Then back substitution occurs as before:
x+λ=0
x=−λ
The solution is therefore x = −λ, y = λ, where λ is any number. There are thus an
infinite number of solutions, forexample
or
x=−1 y=1
x = 1 2
y = − 1 2
and so on.
Observation of the coefficient matrices in the last two examples shows that they have a
determinantofzero.Wheneverthishappensweshallfindtheequationseitherareinconsistentorhaveaninfinitenumberofsolutions.Weshallnextconsiderthegeneralization
ofthismethodtothreeequationsinthreeunknowns.
Example8.36 Solve byGaussian elimination
x−4y−2z =21
2x+y+2z =3
3x+2y−z = −2
Solution We first formthe augmentedmatrix and add the steppedpatternasindicated:
⎛
⎞
1−4−2 21
⎝ 2 1 2 3 ⎠
3 2 −1 −2
The aim is to eliminate all numbers underneath the steps by carrying out appropriate
rowoperations.Thisshouldbecarriedoutbyeliminatingunwantednumbersinthefirst
column first. We find
R 1
R 2
→R 2
−2R 1
R 3
→R 3
−3R 1
⎛
⎝
1−4−2 21
0 9 6−39
0 14 5 −65
We have combined the elimination of unwanted numbers in the first column into one
stage.We now remove unwanted numbers inthe second column:
⎛
⎞
R 1 1−4 −2 21
R 2
⎜ 0 9 6 −39 ⎟
R 3
→R 3
− 14 ⎝
9 R 2 0 0 − 13 ⎠
3 −13 3
and the elimination is complete. AlthoughR 3
→R 3
+ 14 4 R 1
would eliminate the 14, it
would reintroduce a non-zero term into the first column. It is therefore essential to use
the second row and not the first to eliminate this element. We can now read offzsince
⎞
⎠