CRAMER’S RULE
GE 111 – Topic 12

Anton. Section 2.3
1
pp 118-128

GE 111 Engineering Problem Solving

Nov 2014

12. Cramer’s Rule

Learning Outcomes
By the end of this subject the student will be able to:
1. Be able to find the matrix inverse using the Adjoint Method;
2. Solve (e.g. find x, y) a system of linear equations using
Cramer’s Rule.

2

GE 111 Engineering Problem Solving

Nov 2014

Matrix Inversion
How to calculate the matrix inverse?

1
A  adj ( A)
A
1

Where adj (A) is the adjoint matrix.
The adjoint matrix of [A], Adj[A] is obtained by taking the transpose of the cofactor matrix of [A].

adj (A) = [cofactor (A)]T

Anton: 123
3

GE 111 Engineering Problem Solving

Nov 2014

Matrix Inversion: Example 1
1. Minor matrix
2 -4

5

-3

5

7

5

3 -8

2 -4

5

-3

5

7

5

3 -8

2 -4
-3
5

5

M11 =

5

7

3 -8

= -61

5(-8) – 7(3) = -61

M12 =

-3

7

5 -8

= -11

5
7

3 -8

M13 =

-3

5

5

3

= -34

Anton: 122-124; Examples 5 thru 7
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GE 111 Engineering Problem Solving

Nov 2014

Matrix Inversion: Example 1
The resulting matrix of minors is:
-61 -11 -34
[M] =

17 -41

26

-53

-2

29

2. Cofactor matrix, Cij = (-1)i+j Mij
-61

[C] =

11 -34

-17 -41 -26
-53 -29

5

-2

GE 111 Engineering Problem Solving

Nov 2014

Matrix Inversion: Example 1 n 3. Determinant det(A) = |A| =

 a1,j C1,j j=1 |A| = 2(-61) + -4(11) + 5(-34)
|A| = -336

4. Adjoint: adj [A] = [C]T
11 -34

-61 -17 -53

[C] = -17 -41 -26

[C]T = adj [A] = 11 -41 -29

-61

-53 -29

6

-2

GE 111 Engineering Problem Solving

-34 -26

-2

Nov 2014

Matrix Inversion: Example 1
5. Inverse using Adjoint method
A-1

A-1

7

=

1
-336

=

1 det(A) adj(A)

-61 -17 -53

0.182 0.051 0.158

11 -41 -29

= -0.033 0.122 0.086

-34 -26

-2

0.101 0.077 0.006

GE 111 Engineering Problem Solving

Nov 2014

Cramer’s Rule
A system of linear equations can be solved using:
Coefficient matrix
Determinant





ax + by = e

a b

cx + dy = f

det (A) =

8

c d a b c d

x

y

e
=

f

= ad - bc

GE 111 Engineering Problem Solving

Nov 2014

Cramer’s Rule for a 2x2 System
Linear Equations

Coefficient Matrix

Determinant

a b
A=
c d

det (A)

ax + by = e cx + dy = f e b x= f d = det (A)

e f a c a a e c c f
=
y= a det (A) c 9

b d b d =

ed – bf ad – bc

e af – ec f = ad – bc b d

GE 111 Engineering Problem Solving

Nov 2014

Example 2: Solve the System
System of linear equations Coefficient
Matrix

8 5

8x + 5y = 2

A =

2x - 4y = -10

2 -4

Determinant det (A) = 8(-4) – 5(2)
= -32 – 10 = -42

2 5
2(-4) – 5(-10)
42
-10
-4
x=
= 8(-4) – 5(2) = -42 = -1
-42
8 2 y = 2 -10 =
-42
10

8(-10) – 2(2) -84
8(-4) – 5(2) = -42 = 2

GE 111 Engineering Problem Solving

Solution: (-1,2)
Nov 2014

Example 3: Find x and y
System of linear equations Coefficient
Matrix

2 1

2x + y = 1
3x - 2y = -23

A =

3 -2

Determinant det (A) = 2(-2) – 1(3)
= -4 – 3 = -7

1 1
1(-2) – 1(-23) 21
-23
-2 x= = 2(-2) – 1 (3) = -7 = -3
-7

2 1 y = 3 -23
-7
11

=

2(-23) – 1(3) -49
2(-2) – 1(3) = -7 = 7

GE 111 Engineering Problem Solving

Solution: (-3,7)
Nov 2014

Key Points


The denominator consists of the determinant of coefficients matrix (x in the first column, and y in the second column).



The numerator is the same as the denominator, with the constants replacing the coefficients of the variable for which you are solving.

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GE 111 Engineering Problem Solving

Nov 2014

Cramer’s Rule for a 3x3 System
System of linear equations ax + by + cz = j dx + ey + fz = k gx + hy + iz = l

Matrix format a b c x j
A=d e f X= y B= k g h i z l

Determinant a b c det (A) = d e f = aei + bgf + cdh – ceg – bdi - afh g h i
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GE 111