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BA201 Engineering Mathematic
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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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Prepared by : Nur hidayah Othman Page 2
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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Prepared by : Nur hidayah Othman Page 5
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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Prepared by : Nur hidayah Othman Page 8
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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Prepared by : Nur hidayah Othman Page 9
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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Prepared by : Nur hidayah Othman Page 12
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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Prepared by : Nur hidayah Othman Page 13
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
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