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