chapter 11
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Chapter 11
Section 11.4
Solving Larger Systems of Equations
Larger Systems of Equations
A system of equations that has 3 or more variables can be solved by combining both the elimination method and the substitution method.
1. You eliminate variables from the other equations until you get down to one equation with one variable.
2. Solve that one equation for the variable and substitute it back into the other equations and solve for the one variable that remains.
Example Solve the System of Equations:
0993
4452
128124
zyx
zyx
zyx
Multiply Equation 1 by ¼.
0993
4452
323
zyx
zyx
zyx
Multiply Equation 1 by -2 and add to Equation 2.
Multiply Equation 1 by -3 and add to Equation 3.
93
2
323
z
y
zyx
93
2
323
z
y
zyx
Solve Equation 3 for z.
3
2
323
z
y
zyx
Solve Equation 2 for y.
3
2
323
z
y
zyx
3
366
3)3(223
x
x
x
Substitute y and z into Equation 1.
3
2
323
z
y
zyx
The solution is:
x = 3, y = 2, z = -3
Systems of Equations and Matrices
Systems of equations can be represented with matrices in a certain way.
1. Each row corresponds to an equation.
2. Each column to a variable and the last column to the constants.
We write the variables on one side of the equation and the constants on the other. In the matrix separate the variables from the constants with a line (sometimes dashed).
The entries of the matrix are the coefficients of the variables. It is important that if a variable does not show up in an equation that means the coefficient is 0 and that entry in the matrix is 0. The entries on the other side of the line are the constants.
89
743
yx
yx
8
7
19
43
system of equations
Matrix
134
9
853
zy
zx
zyx
system of equations
1
9
8
340
101
531
Matrix
Sometimes algebra might be needed to change the equations to a matrix.
105)3(2
23
yx
xy
10562
23
yx
yx
1652
23
yx
yx
16
2
52
13
Subscripted Variables
In systems of equations where more than 3 variables (sometime for 2 and 3 variables) are needed instead of using regular variable we use just one x but put subscripts on it xi "read x sub i".
6236
8
732
47
0225
431
54321
42
521
5431
xxx
xxxxx
xx
xxx
xxxx
6
8
7
4
0
62306
11111
03020
70011
21205
Matrices with 1's and 0's
A matrix that has 1's down the diagonal from top left to bottom right is easy to read the simultaneous solution of system of equations right from the matrix. They are the entries in the constants column.
74
10
01Matrix
Equations
7
4
y
x
Simultaneous Solution
7,4
8
3
100
010
001
61
8
3
61
z
y
x
8,,3 61
Row Operations
We will use specific row operations to change a matrix from the original equation form to the type with 1's down the diagonal and 0's everywhere else. The row operation we use keep the simultaneous solution
nnnnn
n
n
b
b
b
aaa
aaa
aaa
2
1
21
22221
11211
nc
c
c
2
1
100
010
001
2
1
2221
1211
b
b
aa
aa
ROW OPERATIONS
2
1
10
01
c
c
For a matrix A there are 4 row operations that are allowed. The way you refer to each row operation below the second way is how the TI-83 graphing calculator refers to them.
Ri Rj RowSwap([A],i,j)Interchange the ith and jth rows.
Ri+Rj Row+([A],i,j) Add the ith row to the jth row.
mRi *Row(m,[A],i) Multiply the ith row by the number m.
mRi+Rj *Row+(m,[A],i,j) Multiply the ith row by the number m and add it to the jth row.
RowSwap
Examples of row operations.
3
5
4
211
820
340
A
R1R3
RowSwap([A],1,3)
4
5
3
340
820
211
Row+
3
5
8
212
820
462
A
R1+R3
Row+([A],1,3)
11
5
8
670
820
462
*Row
5
2
4
376
950
16124
A
¼R1
*Row(¼,[A],1)
5
2
1
376
950
431
*Row+
1
5
3
432
820
521
A
-2R1+R3
*Row+(-2,[A],1,3)
5
5
3
610
820
521
Pivoting
The process of using elementary row operations to "clear out" a column and get a 1 in the diagonal position in the column and then zeros in all others is called pivoting. Only the row operations are allowed when doing this process. The diagonal entries are the pivot positions.
nnnn
n
b
b
b
aa
aa
naa
2
1
2
222
112
0
0
1
24
12
93
82A
24
6
93
41
42
6
30
41
½R1
*Row(½,[A],1)
-3R1+R2
*Row+(-3,[Ans],1,2)
The first column is "cleared out"!
42
6
30
41Ans
14
6
10
41
-⅓R2
*Row(-⅓,[Ans],2)
1450
10
01
-4R2+R1
*Row+(-4,[Ans],2,1)
The second column is "cleared out"!
Pivot Position
Gauss-Jordan Elimination
Gauss-Jordan Elimination is a process of pivoting column by column until you have all but the last column "cleared".
24
8
24
20A
R1R2
RowSwap([A],1,2)
8
24
20
24
¼R1
*Row(¼,[Ans],1)
8
6
20
1 21
4
6
10
1 21
-½R2
*Row(-½,[Ans],2)
4
4
10
01-½R2+R1
*Row+(-½,[Ans],2,1)
4
24
1
1220
462
510
A
4
1
24
1220
510
462
R1R2
RowSwap([A],1,2)
½R1
*Row(½,[Ans],1)
4
1
12
1220
510
231
-3R2+R1
*Row+(-3,[Ans],2,1)-2R2+R3
*Row+(-2,[Ans],2,3)
2
1
9
200
510
1301
½R3
*Row(½,[Ans],3)
1
1
9
100
510
1301
13R3+R1
*Row+(13,[Ans],3,1)-5R3+R2
*Row+(-5,[Ans],3,2)
1
4
22
100
010
001
Putting it Together
Lets use matrix simplification (i.e. Gauss-Jordan Elimination) to solve the system of equations below.
752
973
yx
yx
We form the matrix.
7
9
52
73
Call this A and simplify it
7
9
52
73A
7
9
52
73A
⅓R1
*Row(⅓,[A],1)
7
3
52
1 37
-2R1+R2
*Row+(-2,[Ans],1,2)
1
3
0
1
31
37
3R2
*Row(3,[Ans],2)
3
3
10
1 37
3
4
10
01
-7/3R2+R1
*Row+(- 7/3,[Ans],2,1)
Converting back to equations gives:
3
4
y
x
)3,4(Check:
9
2112
)3(7)4(3
7
158
)3(5)4(2
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