BA201 Engineering Mathematic UNIT8 - Cramer's Rule and Inverse Matrix Method

B3001/UNIT8/1

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Prepared by : Nur hidayah Othman Page 1

Unit

8

CRAMER’S RULE AND

INVERSE MATRIX METHOD

To know the different types of matrices and

understand how to apply it on simple algebra problem

solving.

.

Upon completing this module, you should be able to:

Obtain linear equation solution using inverse

matrix method.

Obtain linear equation solution using Cramer’s

Rule.

General Objectives

Specific Objectives

B3001/UNIT8/2

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8.0 INTRODUCTION

After understanding how to solve the matrices operation, we should be able to solve linear

equation using Cramer’s Rule and Inverse Matrices Method.

8.1 INVERSE MATRIX METHOD

Consider one set of three equations:

1131211 bzayaxa

2232221 bzayaxa

3333231 bzayaxa

Then, write the equation in a matrix from;

333231

232221

131211

aaa

aaa

aaa

z

y

x

=

3

2

1

b

b

b

If we take the matrix

333231

232221

131211

aaa

aaa

aaa

as matrix A and matrix

z

y

x

as c and matrix

3

2

1

b

b

b

as b,

We can write the above matrix equation as:

Ac = b

And multiply both equation parts with A-1

A-1

Ac = A-1

b

INPUT

B3001/UNIT8/3

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A-1

A will become identity matrix, I.

c = A-1

b

To obtaine c =

z

y

x

we need to multiply inverse of A with b. Therefore, the key to solve this

problem is get the inverse A, which is A-1

.

Example 8.1:

Determine the solution for the set of linear equation below:

x + 3y + 3z = 4

2x –3y –2z = 2

3x + y + 2z = 5

Solution:

The first step is write in form matrix equation,

z

y

x

=

5

2

4

Take matrix as A, determine the inverse A,

First of all, determine determinant A , A = -1 and minor A, which is

983

873

11104

213

232

331

213

232

331

B3001/UNIT8/4

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Then, determine co-factor of A, which is

983

873

11104

Next, determine ad joint A, which is

9811

8710

334

Therefore, inverse of A is

9811

8710

334

Then,

z

y

x

=

9811

8710

334

5

2

4

z

y

x

=

15

14

7

Therefore, the answer x = 7, y = 14, z = -15

B3001/UNIT8/5

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Example 8.2:

Define the solution for the set linear equation below:

3x + 2y - z = 10

7x – y + 6z = 8

3x + 2z = 5

Solution:

As in example 7.1 above,

Write the equation in matrix form:

203

617

123

z

y

x

=

5

8

10

Determine the inverse of matrix

203

617

123

, which is,

1763

2594

1142

And solve the matrix equation:

z

y

x

=

5

8

10

1763

2594

1142

z

y

x

=

7

13

3

Therefore, the answer is x = -3, y = 13, dan z = 7

B3001/UNIT8/6

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ACTIVITY 8a

8a.1 By using the inverse matrix method, solve the following linear equations:

a) 2x1 -x2 +3x3 =2

x1 +3x2 -x3 =11

2x1 -2x2 + 5x3 =3

b) x1 + 3x2 + 2x3 = 3

2x1 - x2 - 3x3 = -8

5x1 + 2x2 + x3 = 9

B3001/UNIT8/7

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FEEDBACK 8a

8a.1 a) x1=2

x2=5

x3=3

b) x1=2

x2=-3

x3=5

B3001/UNIT8/8

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8.2 CRAMER’S RULE

Another one method to solve the linear equation using matrix is using the Cramer’s Rule.

Cramer’s Rule needs skill to obtained determinant in a matrix.

If we get the matrix equation until here,

333231

232221

131211

aaa

aaa

aaa

3

2

1

x

x

x

=

3

2

1

b

b

b

By take A =

333231

232221

131211

aaa

aaa

aaa

, we can obtained 1x = A

A1

Where, 1A =

33323

23222

13121

aab

aab

aab

See the column

3

2

1

b

b

b

replace to column

31

21

11

a

a

a

in A to obtaine 2x = A

A 2.

Where, 2A =

33331

23221

13111

aba

aba

aba

which the column

3

2

1

b

b

b

replace to column

32

22

12

a

a

a

in A

INPUT

B3001/UNIT8/9

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And to obtaine 3x = A

A3

Where, 3A =

33231

22221

11211

baa

baa

baa

and here column

3

2

1

b

b

b

replace to column

33

23

13

a

a

a

in A.

Example 8.3:

Solve the following linear equation:

5x - y + 7z = 4

6x - 2y + 9z= 5

2x + 8y –4z= 8

Solution:

Write in matrix equation form:

482

926

715

8

5

4

z

y

x

By take A =

482

926

715

A1 =

488

925

714

A2 =

482

956

745

and A3 =

882

526

415

A =2 1A = 44 2A = -26 dan 3A = -34

x = A

A1 =

2

44= 22 y =

A

A 2 =

2

26= 13 z =

A

A 3 =

2

34= -17

B3001/UNIT8/10

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Example 8.4:

Solve the following linear equation:

2x -3y + z = 5

x + y + z = 7

3x – 4z = 10

Solution:

Write in matrix equation form:

403

111

132

10

7

5

z

y

x

By take A =

403

111

132

A1 =

4010

117

135

A2 =

4103

171

152

and A3 =

1003

711

532

A =2 1A = 44 2A = -26 dan 3A = -34

x = A

A1 =

2

44= 22 y =

A

A 2 =

2

26= 13 z =

A

A 3 =

2

34= -17

B3001/UNIT8/11

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ACTIVITY 8b

8b.1 Solve the following linear equation by using the Cramer’s Rule:

a) X + 2y - 3z = 3

2x – y – z = 11

3x + 2y + z = -5

b) X - 4y - 2z = 21

2x + y + 2z = 3

3x+2y-z=-2

B3001/UNIT8/12

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FEEDBACK 8b

8b.1 a) x = 2

y = -4

z = -3

b) x = 3

y = -5

z = 1

B3001/UNIT8/13

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SELF ASSESMENT

8.1 Solve the following linear equation by using the inverse matrix method:

a) 2i +j + k = 8

5i – 3j + 2k =3

7i + j +3k = 20

b) 3x + 2y + 4z = 3

x + y + z = 2

2x – y + 3z = -3

8.2 Solve the following linear equation by using the Cramer’s Rule.

a) 4a – 5b + 6c = 3

8a – 7b – 3c = 9

7a – 8b + 9c = 6

b) 3x + 2y – 2z = 16

4x + 3y + 3z = 2

-2x + y + z = 1

B3001/UNIT8/14

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FEEDBACK

8.1 a) I = 2

j =3

k =1

b) x = 1

y = 2

z = -1

8.2 a) a = 2

b = 1

c = 0

b) x = 2

y =1.5

z = -3.5

Documents

BA201 Engineering Mathematic UNIT8 - Cramer's Rule and Inverse Matrix Method