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

290 Chapter 8 Matrix algebra
thelastequationstates0x +0y − 13 3 z = −13 ,thatisz = 1.Backsubstitutionofz = 1
3
in the second equation gives y = −5 and, finally, substitution of z and y into the first
equation givesx = 3.
Example8.37 Solve the following equationsby Gaussian elimination:
x−y+z =3
x+5y−5z =2
2x+y−z =1
Solution Forming the augmented matrix,wefind
⎛ ⎞
1−1 13
⎝ 1 5−52⎠
2 1 −1 1
Then, as before, we aim to eliminate all non-zero entries under the step. Starting with
those inthe first column, wefind
⎛
R 1
R 2
→R 2
−R 1
⎝ 1 −1 1 3
⎞
0 6−6−1⎠
R 3
→R 3
−2R 1 0 3 −3 −5
Then,
R 1
R 2
R 3
→2R 3
−R 2
⎛
⎝ 1 −1 1 3
0 6−6−1
0 0 0 −9
Thislastlineimpliesthat0x+0y+0z = −9,whichisclearlyinconsistent.Weconclude
that thereareno solutions.
⎞
⎠
You will see from Examples 8.36 and 8.37 that not only have all entries under the step
beenreducedtozero,butalsoeachsuccessiverowcontainsmoreleadingzerosthanthe
previousone.Wesaythesystemhasbeenreducedtoechelonform.Moregenerallythe
systemhasbeenreducedtoechelonformiffori < jthenumberofleadingzerosinrow
j islarger thanthe number inrowi. Consider Example 8.38.
Example8.38 Solve the following equationsby Gaussian elimination:
2x−y+z =2
−2x+y+z =4
6x−3y−2z = −9
Solution Forming the augmented matrix,wehave
⎛
⎝ 2 −1 1 2
⎞
−2 1 1 4 ⎠
6 −3 −2 −9