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SOLVING SYSTEM OF LINEAR EQUATIONS Method : Gaussian Elimination The Gaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix A with the number 1 as the entry down the main diagonal and have all zeros below.
The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below. IMPORTANT NOTE: WE WILL FOCUS ON GETTING OUT MATRIX IN REDUCED ROW ECHELON FORM WHICH WILL LOOK LIKE THIS;
How To Given an augmented matrix, perform row operations to achieve row-echelon form.
- The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary.
- Use row operations to obtain zeros down the first column below the first entry of 1.
- Use row operations to obtain a 1 in row 2, column 2.
- (^) Use row operations to obtain zeros down column 2, below the entry of 1.
- Use row operations to obtain a 1 in row 3, column 3.
Example 1 Writing an Augmented Matrix Write the augmented matrix for the system of linear equations.
23 xx^25 yy 6 zz 33
x 3 y 4 z 2
-->
33
2 (^16) 4 52
3 23
1
Linear System Augmented Matrix Exercises :Write the Augmented for the system of linear equations Linear system Augmented Matrix a) 3 xx^45 yyz 9 z^38
x 3 y 4 z 7 8
b) x 9 yx 4 yz^36
3 x 6 y 4 z 7
6
c) 3 xx^28 yy^4 zz 16
2 y 5 z 3 1
d) 4 x 3 yx (^2) zz 14 2 y 7 z 11 1
e) 7 y 3 x (^2) yz (^41) 2 x 7 z 11 1
f) 3 xx y 2 y 6 z 6 z 8 w 7 w (^12) 72 xx 23 yy 67 zz 53 ww 411
1 2 6 7 2
Example 3: Gaussian Elimination and back substitution Solve the following system of equation
32 xx 64 yy 53 zz 01
x y 2 z 9
Solution: Write the system of equations to augmented matrix.
- 3R R
- 2R R
1 3
1 2
- 17/2^9
21 R^20111 - 7/2^2
- 17/2^9
01 11 - 7/2^2
- 3R 2 R 3
- 2R 3
The system corresponding to this matrix is
z 3 2
y 27 z^17
x y 2 z 9
Back Substitute to solve the system
x 9 y 2 z 9 ( 2 ) 2 ( 3 ) 1
y^17227 ( 3 ) 2
Z 3
02 24 73 171
2 2 4 18
Change to 0
Change to 1
03 26 75 017
3 3 6 18
2 R 1 R 2
Change to 0
X (-3)
3 R 1 R 3
00 03 117 1727
0 3 21 / 2 51 / 2
Change to 1
X( -2) X (-3)
3 R 2 R 3
Example 4: Solve the following system of equation
3 xx 72 yy 43 zz 101
x y 2 z 8
Solution: The augmented matrix of the system is
We perform elemental operations in the rows to obtain the reduced rowechelon form
3 R R
R R
1 3
1 2
10R 2 R 3
521 R^2
The system corresponding to this matrix
z 2
y-5z -
x y 2z 8
Back Substitute to solve the system
x 8 y 2 z 8 1 2 ( 2 ) 3
y 9 5 z 9 5 ( 2 ) 1
Z 2
Solution: x=3 ,y=1, z=
Example 6 Solve the following system of equation
3 x y 2 z 11
x 2 y z 8
2 x y 2 z 10
Solution:
11
R 1 R 2 21 - - 1211
00 53 -^01
32 RR 11 RR 32
-^82
31 R^201 -^1210
00 01 -^01
5 R 2 R 3
-^82
R 3
The system corresponding to this matrix
z 3
y -
x 2 y z 8
Back Substitute to solve the system
x 8 2 y z 8 2 ( 2 ) 3 1
y 2
Z 3
Solution: x 1 ,y 2 ,z 3
Step 1 accomplished by interchanging row 1 and row 2:We want a 1 in row 1, column 1.This can be
Step 2 : Now let’s obtain a 0 in row 2 and Row 3 (column 1)
Example 7 Solve the following system of equation
y 6 z 1
x 5 z 7
x 2 y 3 z 15
Solution:
00 - 12 -^26
R 1 R 2
-^41
21 R^1
01 12 -^31
R 1 R 2
00 01 1 -^1
51 R^2
The system corresponding to this matrix
z 1
y z 4
x 2 y 3 z 15
Back Substitute to solve the system
xy 154 z 2 y^431 z^5152 ( 5 ) 3 ( 1 ) 2
Z 1
Solution: x 2 ,y 5 ,z 1
