SLIDE DBM10063 CHAPTER 5

DBM10063
CHAPTER 5

MATRICES AND LINEAR
ALGEBRA

5.0 MATRIX
& LINEAR ALGEBRA

5.1 UNDERSTAND MATRICES

ROW COLUMN

• Horizontal line • Vertical line

235 235
•103 •103

625 625

6 21 This matrix is 2×3
−1 9 5 (2 rows by 3 columns)

ELEMENTS OF A MATRIX ORDER OF A MATRIX
(mrow x ncolumn)
235
103 235 3x 3 matrix
625 103
625
Element : All the number in a matrix
12 2 x 2 matrix
34

8 3x1 matrix
1 1 x 3 matrix
3

425

5.1.2 TYPES OF MATRIX

1. Square Matrix – no of rows and column is equal 2. Zero Matrix- all elements is zero
12
34
0
235
103
625

3. Diagonal Matrix – all elements of a square 4. Identity Matrix- elements of a diagonal matrix
matrix is zero except diagonal. consist value 1

200 100
010 I= 0 1 0
005
001

IDENTITY MATRIX

• AI=A

• 1 2 1 0 = 1 2
3 4 0 1 3 4

• IA=A

• 1 0 1 2 = 1 2
0 1 3 4 3 4

TRANSPOSITION OF A MATRIX

11 12 = 11 21 31
22 12 22 32
• IF = 21 32
31

EXERCISE 1 = 2 3 5
21 1 −2 2

= 3 −2 147
52 = 2 5 8

EXERCISE 2 369

123
= 4 5 6

789

OPERATION ADDITION
OF SUBTRACTION
MULTIPLICATION
MATRICES

A) ADDITION

1 2 + 3 2 = 4 4
3 4 4 1 7 5

132 235 367
2 0 4+5 1 1=7 1 5
421 370 791

B) SUBSTRACTION

Example 1:

1 2 − 3 2 = −2 0
3 4 4 1 −1 3

Example 2:

1 3 2 2 3 5 −1 0 −3
2 0 4 − 5 1 1 = −3 −1 3
4 2 1 3 7 0 1 −5 1

C) MULTIPLICATION

Example 1:

2 3 2 = 6 4 1 3 2 2 3 5 23 20 8
4 1 8 2 2 0 4 5 1 1=
421370
1 3 2 2 3 5 23
2 0 4 5 1 1= Solution:
421370 1×5 + 3×1 + 2×0 =8

Solution: 1 3 2 2 3 5 23 20 8
2 0 4 5 1 1 = 16
1 × 2 + 3 × 5 + 2 × 3 = 23 421370

1 3 2 2 3 5 23 20 Solution:
2 0 4 5 1 1= 2 × 2 + 0 × 5 + 4 × 3 = 16
421370

Solution:
1 × 3 + 3 × 1 + 2 × 7 = 20

1 3 2 2 3 5 23 20 8 1 3 2 2 3 5 23 20 8
2 0 4 5 1 1 = 16 34 2 0 4 5 1 1 = 16 34 10
421370 4 2 1 3 7 0 21 21

Solution: Solution:
2 × 3 + 0 × 1 + 4 × 7 =34 4 × 3 + 2 × 1 + 1 × 7 = 21

1 3 2 2 3 5 23 20 8 1 3 2 2 3 5 23 20 8
2 0 4 5 1 1 = 16 34 10 2 0 4 5 1 1 = 16 34 10
421370 4 2 1 3 7 0 21 21 22

Solution: Solution:
2 × 5 + 0 × 1 + 4 × 0 = 10 4 × 5 + 2 × 1 + 1 × 0 = 22

1 3 2 2 3 5 23 20 8
2 0 4 5 1 1 = 16 34 10
4 2 1 3 7 0 21

Solution:
4 × 2 + 2 × 5 + 1 × 3 = 21

In General:

To multiply an m×n matrix by an n×p matrix, the n must be the same,
and the result is an m×p matrix.

Example 2:

1 2 3 × 7 8 = 58
4 5 6 9 10
11
12

Solution:

1 × 7 + 2 × 9 + 3 × 11 = 58

1 2 3 7 8 58 64 Solve by
4 5 6 9 10 ? ? yourself
11 12
× =

Solution:

1 × 8 + 2 × 10 + 3 × 12 = 64

Example 3:

row 4 3 1 3 5
0 1 7 0 9 =undefined
−3 2 5 3 1

If elements in ROW is
not same as the number
C of COLUMN of second
o matrix= NOT DEFINED
l
u
m
n

ACTIVITY 1

1. Express the following matrices

a) −5 9 + 4 0
−6 5 2 −8

b) 3 + −1
−9 8

0 −3 6 28 2
c) 6 0 4 + 1 7 −4

2 2 −19 3 12 0

2. Based on the following matrices

A= 3 7 B = 3 4 C= 4 5 D= 4 −2
9 5 −3 −5 2 −1 5 7

a) A + B

b) A – C

c) D + ( B – A)

d) B + C

2 15
3. D = −9 = 3 . Find D+E

7 −2

21 05
4. C = 3 −5 = −4 5 . Find:

69 63

a) C+B

b) C-B

3 5 −3 1 3 5 R= 2 0 3 S= 4 2
8 4 Q= 7 9 7 −1 4 9 6 1 . Find:
5. P= 0 7 2 3 1 3 5
5
−1

a) PQ

b) 2

c) QI

d) QR

e) RS

f) SQ

DETERMINANT OF MATRICES

A) Determinant of 2x2 matrix

Example 1 : If = 1 2 . Find det(A)
3 4

Solution: = 1 2
3 4

= 1×4 − 3×2
= −2

B) Determinant of 3x3 matrix

1 −2 0
Example 2 : If B = 4 7 −1 . Find det(B)

23 5

Solution: = 1 7 −1 − −2 4 −1 +0 4 7
3 5 2 5 2 3

= 1ሾ 7 × 5 − −1 × 3 ሿ − (−2)ൣ 4 × 5 − −1 × 2 ሿ + 0ሾ 4 × 3 − 7 × 2 ሿ

= 1 35 + 3 + 2 20 + 2 + 0
= 38 + 44
= 88

ACTIVITY 2

1. Find the determinant for the matrix below:

5 −2 3
= 4 −1 −5

13 1

911
= 4 2 3

163

1 −5 3 2 35
2. Given = −2 5 7 = 2 0 8

9 −6 5 −6 5 7
a) Determinant

b) −

5.2 EXPLAIN SYSTEMS OF LINEAR EQUATIONS

1. Cramer’s Rule Linear equations up to
three variables using
2. Inverse Matrix

Method

Cramer’s Rule

1.Write in a matrix form =

2. Find the determinant of ,

3. Find 1 by substituting ‘b’ into column 1 of A. Calculate 1

4. Find 2 by substituting ‘b’ into column 2 of A. Calculate 2

5. Find 3 by substituting ‘b’ into column 3 of A. Calculate 3
6. Find three variable using formula:
= 1 , = 2 , z= 3



EXAMPLE 1: SOLVE THE SIMULTANEOUS LINEAR EQUATION BELOW USING
CRAMER’S RULE

4 − + =
− − =
7 − + =

1.Write in a matrix form =

4 −5 6 3

8 −7 −3 = 9 Click to add text

7 −8 9 6

2. Find the determinant of ,

4 −5 6
= 8 −7 −3

7 −8 9

=4 −7 −3 − (−5) 8 −3 +6 8 −7
−8 9 7 9 7 −8
= 4 −63 − 24 + 5 72 + 21 + 6 −64 + 49

= −348 + 465 − 90

= 27

3. Find 1 by substituting ‘b’ into column 1 of A. Calculate 1

3 −5 6 1 =3 −7 −3 − −5 9 −3 +6 9 −7
1 = 9 −7 −3 −8 9 6 9 6 −8
= 3 −63 − 24 + 5 81 + 18 + 6 −72 + 42
6 −8 9
= −261 + 495 − 180

= 54

Click to add text

4. Find 2 by substituting ‘b’ into column 2 of A. Calculate 2

43 6 2 =4 9 −3 − 3 8 −3 +6 8 9
6 9 7 9 7 6
2 = 8 9 −3 = 4 81 + 18 − 3 72 + 21 + 6 48 − 63
76 9
= 396 − 279 − 90

= 27

5. Find 3 by substituting ‘b’ into column 3 of A. Calculate 3

3 = 4 −5 3 3 =4 −7 9 − −5 8 9 +3 8 −7
8 −7 9 −8 6 7 6 7 −8
7 −8 6 = 4 −42 + 72 + 5 48 − 63 + 3 −64 + 49

= 120 − 75 − 45

=0

6. Find three variable using formula:
= 1 , = 2 , z= 3



= 54 , = 27 , z = 0

27 27 27

= 2; = 1; = 0;

EXERCISE1 : SOLVE THE SIMULTANEOUS LINEAR EQUATION BELOW USING CRAMER’S RULE

a) + 2 − = 4 b) 2 − 5 + = 3 c) + + = 12
3 − 4 − 2 = 2 − 2 − 2 = 5 + 2 + 3 = 6
5 + 3 + 5 = −1 3 − + 3 = 2 2 − − 4 = −3

Exercise2 : the simultaneous linear equation is given below

− − = 2
4 + 2 + 3 = 7
3 − 5 + 6 = −5

a) Write in the matrix form
b) Find the determinant of the matrix
c) Find the value of Z using Cramer’s rule

Inverse Matrix Method

1.Write in a matrix form =
2. Find the determinant of A
3. Find the minor of A
4. . Find the cofactor of A
5. Find the adjoint of A
6. Find the inverse of A, −1

EXAMPLE 1: SOLVE THE SIMULTANEOUS LINEAR EQUATION BELOW USING INVERSE MATRIX
METHOD

+ + =
+ + =

− + = −

Solution:

1.Write in a matrix form =

3 2 4 3

1 1 1 = 2

2 −1 3 −3

2. Find the determinant of A

324
= 1 1 1

2 −1 3

=3 1 1 −2 1 1 +4 1 1
−1 3 2 3 2 −1
= 3 3 + 1 − 2 3 − 2 + 4 −1 − 2

= −2

3. Find the minor of A

324 11 12 13
= 1 1 1 = 21 22 23

2 −1 3 31 32 33

11 = 1 1 =3− −1 =4
−1 3

12 = 1 1 =3−2=1 31 = 2 4 = 2 − 4 = −2
2 3 1 1

13 = 1 1 = −1 − 2 = −3 32 = 3 4 = 3 − 4 = −1
2 −1 1 1

21 = 2 4 =6− −4 = 10 33 = 3 2 =3−2=1
−1 3 1 1

22 = 3 4 =9−8=1
2 3
4 1 −3
23 = 3 2 = −3 − 4 = −7 , = 10 1 −7
2 −1
−2 −1 1

4. . Find the cofactor of A Change this 4 element
sign
4 1 −3
= 10 1 −7 4 −1 −3
Cofactor = −10 1 7
−2 −1 1
−2 1 1

5. Find the adjoint of A 4 −10 −2
Adjoint = −1 1 1
4 −1 −3
Cofactor = −10 1 7 −3 7 1

−2 1 1

6. Find the inverse of A, −1

Formula: −1 = 1

4 −10 −2
−1 = 1 4 −10 −2 −2 −2 −2
−2 −1 1 1 −1 = −1 1 1
−3 7 1 −2 −2 −2
−3 7 1
−2 −2 −2

−2 5 1
11 1
−1 = 2 −2 − 2
3 71
2 −2 −2

EXERCISE

302
1. Given = 1 −1 3 . Find the cofactor of A.

024

3 12
2. Given = −1 2 1

4 15

a) calculate the determinant of B
b) Find the minor of B

3. Solve the following simultaneous linear equation by using the Inverse Matrix

2 + − = 8
2 + 2 = 11
3 + = 10