F UNDAMENTALS OF E NGINEERING
A NALYSIS
Eng. Hassan S. Migdadi


Evaluate the determinant of a 3 x 3 matrix, and

Find the area of a triangle given the coordinates of its vertices.


Every square matrix has a determinant. The determinant has the same elements as the matrix, but they are enclosed between vertical bars instead of brackets. you have learned a method for evaluating a 2 x 2 determinant.


The determinant of


3
17

11
2


3
 is
17
2
11
. To evaluate
The determinant, use the rule for second order determinants.
17

3

2
11

3 ( 11 )

(

2 )( 17 ) a c
33

(

34 ) or 67 d b
 ad
 bc

Expansion by Minors

A method called expansion by minors can be used to evaluate the determinant of a 3 x 3 matrix. The minor of an element is the determinant formed when the row and column containing that element are deleted. For the determinant
1
4
9
3
8
5
7 1
2 , the min orof 5 is 4
6 9 the min orof 1 is

1
4
9
3
8
5
7
2 or
8
5
6
3
8
5
7
2 or
1
4
6
2
6
7
2
.

Expansion by Minors

To use expansion by minors with third-order determinants, each member of one row is multiplied by its minor. The signs of the products alternate, beginning with the second product. The definition below shows an expansion using the elements in the first for of the determinant. However, ANY row can be used. d g a e h b i f
Expansion of a Third-Order Determinant c
 a h e f i
 b d g f i
 c d g e h

a d g
Ex. 1: Evaluate the determinant of
Using expansion by minors.
b e h i f c
 e a h i f
 b d g i f
 c d g e h


2


6
1
2
6
3
5
4
7
8

2
5
2
7
8

6
3
1
7
8

6
4
1
1 2

2 ( 40

14 )

3 ( 48

7 )

4 ( 12

5 )

52

123

28
 
43
5
2
3
5
2
4
7


8



Using Diagonals


Another method for evaluating a third order determinant is using diagonals.
STEP 1: You begin by repeating the first two columns on the right side of the determinant.
a d g b e h i f c
 d a g b e h i f c a d
 e b g h

Using Diagonals

STEP 2: Draw a diagonal from each element in the top row diagonally downward. Find the product of the numbers on each diagonal.
a d g b e h i f c
 d a g b e h f c a d
 e b i g
h

Using Diagonals

STEP 3: Then draw a diagonal from each element in the bottom row diagonally upward. Find the product of the numbers on each .
a d b e c f
 d a b e f c gec hfa a b d
 e g h i g h i g h idb

Using Diagonals

To find the value of the determinant, add the products in the first set of diagonals, and then subtract the products from the second set of diagonals.
The value is: aei + bfg + cdh – gec – hfa – idb

Ex. 2: Evaluate using diagonals.






3
1
3
4

2

1

0
2
5




First, rewrite the first two columns along side the determinant.

1 4 0

1 4

3
3


2
1

5
2
3

3
 
2

1

Ex. 2: Evaluate using diagonals.
Next, find the values using the diagonals.
0

1 4 0

1
-5
3

2

5 3






3
1
3
4
 
2
4

2

1
24

0
2
5





3

1 2
4

3
60

1
0
Now add the bottom products and subtract the top products.
4 + 60 + 0 – 0 – (-5) – 24 = 45. The value of the determinant is 45.

Area of a triangle

Determinants can be used to find the area of a triangle when you know the coordinates of the three vertices. The area of a triangle whose vertices have coordinates (a, b), (c, d), (e, f) can be found by using the formula:
A

1
2 a e c b d f 1
1
1 ,

Ex. 3: Find the area of the triangle whose vertices have coordinates (-4, -1), (3, 2), (4, 6).
How to start: Assign values to a, b, c, d, e, and f and substitute them into the area formula and evaluate.
a = -4, b = -1, c = 3, d = 2, e = 4, f = 6
A

1
2 a c e b d f 
4

1 1
 8 -24
-3
4

1
A

2
1
4
3
6
2
1
1
4
3
-8 -4

6
2
18
Now add the bottom products and subtract the top products.
-8 + (-4) + 18 – 8 – (-24) –(-3) = 25. The value of the determinant is 25. Applied to the area formula ½ (25) = 12.5. The area of the triangle is 12.5 square units.
1
1 ,
1

Sometimes one or more of the elements of a determinant may be unknown, but the value of the determinant is known. You can use expansion to find the values of the variable.
Ex. 4: Solve for n if
4
2

4
3
2 n

3 n
6 4
4
2

4
3
7
3
2

4

2 n

3 n
3
2 n

3 n
6
7
 
582
3
24n – 84 – 36n + 48n +84n – 18 = -582
120n – 102 = -582
Collect like terms
120n = -480 n = -4
Add 102 to both sides
Divide by 120 both sides.
The value of n is -4.