Semester 1 Mathematics STPM (Student's Edition)

146

Method 2 + 4 2

3 1
∑ ( + 1)( + 2) = ∑ [ − + 1 + + 2]
=1 =1

= ∑ =1 [(2 − 2) − (1 − +1 2)]

+1 +1

= ∑ =1 [2 − +21] − ∑ =1 [ +11 − +12]

Let ( ) = 1 , Let ( ) = 1
+1


+ 4

∑ ( + 1)( + 2) = 2 ∑[ ( ) − ( + 1)] − ∑[ ( ) − ( + 1)]
=1 =1 =1

= 2[ (1) − ( + 1)] − [ (1) − ( + 1)]

= 2 [11 − +1 1] − [21 − +1 2]

=3− 2 + 1

2 +1 +2

= 3 − 2( +2)−( +1)
2 ( +1)( +2)

= 3 − +3
2 ( +1)( +2)

As ⟶ ∞, 1 → 0, 1 → 0 and the summation approaches 3.
+1 +2
Therefore, the series is convergent with ∞ = 32. 2

EXAMPLE 4

Given that is a positive integer and ( ) = 1 simplify ( ) − ( + 1) .
2
3 5 7
Hence, find the sum of the first terms of the series 12×22 + 22×32 + 32×42 + ⋯ .

Solution:

( ) − ( + 1)
11

= 2 − ( + 1)2
( + 1)2 − 2

= 2( + 2)2
2 + 2 + 1 − 2

= 2( + 1)2
2 + 1

= 2( + 1)2

12 3 22 + 22 5 32 + 32 7 42 + ⋯ + ℎ
× × ×

147

2 + 1
= ∑ 2( + 1)2

=1


11
= ∑ [ 2 − ( + 1)2]

=1


= ∑[ ( ) − ( + 1)]

=1

= (1) − ( + 1)
11

= 12 − ( + 1)2
1

= 1 − ( + 1)2
( + 1)2 − 1

= ( + 1)2
2 + 2

= ( + 1)2
( + 2)

= ( + 1)2

EXAMPLE 5

Express 1 in partial fraction.
(3 −2)(3 +1)

Hence, find ∑ =1 1 and deduce ∑∞ =1 1 .
(3 −2)(3 +1) (3 −2)(3 +1)

Solution:

Let 1 ≡ +
(3 −2)(3 +1) (3 −2) (3 +1)

1 = (3 + 1) + (3 − 2)

Let = − 1 ⇒ = − 1
3
3

Let = 2 ⇒ = 1

33

Therefore, 1 ≡ 1 − 1
(3 −2)(3 +1) 3(3 −2) 3(3 +1)

1 1 1
∑ (3 − 2)(3 + 1) = ∑ [3(3 − 2) − 3(3 + 1)]
=1 =1

=1−1 = 1

3(1) 3(4)

+1 − 1 = 2
= 3
3(4) 3(7)

+ 1 − 1
3(7) 3(10)

⋮⋮

148

+ 1 − 1 =
3(3 −2) 3(3 +1)

11
= 3 − 3(3 + 1)

(3 + 1) − 1
= 3(3 + 1)

3
= 3(3 + 1)


= 3 + 1

∞1 lim 11
∑ = (3 − 1))
(3 − 2)(3 + 1) →∞ 3(3 +
=1

1
=3−0

1
=3

EXERCISE 2.4

1. a) Express 2 in partial fractions.
(2 +1)(2 +3)

b) Show that ∑ =1 2 = 1− 1 .
(2 +1)(2 +3)
3 2 +3

c) Find the sum of the series 1 +1 + 1 + 1 + ⋯+ 1 and find ∑∞ =1 1 .
3×5 7×9 9×10 37×39 (2 +1)(2 +3)
5×7

2. If ( ) = 1 , obtain and simplify ( + 1) − ( ) . Hence, find ∑ =1 1 .
( +1) ( +1)( +2)

3. Show that if ( ) = 1 then ( ) − ( − 1) = −4 . Hence, or otherwise,
(2 +1)(2 +3) (2 −1)(2 +1)(2 +3)

find the sum of the series 1 + 1 + 1 + 1 + ⋯ + 1 .
1×3×5 3×5×7 5×7×9 7×9×11 (2 −1)(2 +1)(2 +3)

149

STPM PAST YEAR

STPM MM 2013 (Section B)

Q1 a) Express 1 in partial fractions, and deduce that
( 2−1)

1 ≡ 1 [ ( 1−1) − ( 1+1)] . [4 marks]
( 2−1) 2

Hence, use the method of differences to find the sum of the first ( − 1) terms, −1 , of the

series

11 1 1
2 × 3 + 3 × 8 + 4 × 15 + ⋯ + ( 2 − 1) + ⋯ ,

and deduce . [6 marks]

b) Explain why the series converges to 1 , and determine the smallest value of such that
4
1
4 − < 0.0025 . [5 marks]

ANSWER 1(a) 2 11
Section ∴ (2 + 1)(2 + 3) ≡ 2 + 1 − 2 + 3
2.2

Exercise
2.4

1(b) 2
1(c) 13

2 ∞1 1
∑ (2 + 1)(2 + 3) = 6
=1

2
( + 1) − ( ) = − ( + 1)( + 2)

( + 3)

4( + 1)( + 2)

3 2 + 2
3(2 + 1)(2 + 3)

150

ANSWER 1(a) 11 1
Section ∴ 2 − 1 ≡ 2( − 1) − 2( + 1)
2.2

Exercise 11 1
2.4 −1 = 2 [2 − [( − 1) + 1][( − 1) + 2]]

STPM 11 1
Pass Year ∴ = 2 [2 − ( + 1)( + 2)]

(b) = 13

2.3 BINOMIAL Learning outcome:
EXPANSIONS a) Expand (a + b)n, where n ϵ Z+.
b) Expand (1 + x)n, where n ϵ Q, and identify the condition |x| < 1 for the validity of

this expansion.

c) Use binomial expansions in approximations.

Learning outcome: Expand (a + b)n, where n ϵ Z+.

n! = n(n - 1)(n - 2)… 3⸱2⸱1, with n ϵ ℤ+,

0! = 1

(n + 1)! = (n + 1)⸱n!

The binomial coefficient, ( ) = ! ! for n, r ϵ ℤ+ and 0 ≤ r ≤ n .
( − )!



Binomial Theorem, ( + ) = ∑ =0( ) − , n ϵ ℤ+.

( + ) = + ( 1 ) −1 + ( 2 ) −2 2 + ( 3 ) −3 3 + … + ( ) − + ⋯ + ( 1) −1 +
−

EXAMPLE 1
Using the binomial theorem, find the expansion of (2x + 3y)4.

Solution:
(2 + 3 )4 = (2 )4 + (41) (2 )3(3 ) + (24) (2 )2(3 )2 + (34) (2 )(3 )3 + (3 )4

= 16 4 + 96 3 + 216 2 2 + 216 3 + 81 4

EXAMPLE 2

Using the binomial theorem, find the expansion of (a + b)4 .

Hence, find

(a) (2x + 3y)4 (b) ( − 2)4

2

151

Solution:
( + )4 = ( )4 + (14) ( )3( ) + (24) ( )2( )2 + (43) ( )( )3 + ( )4

= 4 + 4 3 + 6 2 2 + 4 3 + 4

(a) (2x + 3y)4 = (2 )4 + 4(2 )3(3 ) + 6(2 )2(3 )2 + 4(2 )(3 )3 + (3 )4
= 16 4 + 96 3 + 216 2 2 + 216 3 + 81 4

(b) ( − 2)4 = ( )4 + 4 ( )3 (− 2) + 6 ( )2 (− 2)2 + 4 ( ) (− 2)3 + (− 2)4
2 22 2 2

= 4 − 3 2 + 3 2 4 − 2 6 + 8
16 2 2

EXAMPLE 3
Find the coefficient of x6 in the expansion of (3x2 + 4)7.

Solution:

( + ) = ∑ =0( ) −
where the (r + 1)th term = ( ) −

7

Thus (3 2 + 4)7 = ∑ =0(7 )(3 2)7− (4)
The term in x6 is when 2(7 – r) = 6, r = 4

Coefficient of x6 = (47) 3344 = 241920

The Expansion of (1 + x)n , n ϵ ℤ+.

(1 + ) = 1 + ( 1 ) + ( 2 ) 2 + ( 3 ) 3 + … + ( ) + ⋯ + ( 1) −1 +
−

= 1 + + ( −1) 2 + ⋯ + ( −1)… ( − +1) + ⋯ + −1 +

2! !

EXAMPLE 1
Use the binomial theorem to expand (1 + x)5.

Solution:

(1 + )5 = 1 + 5 + 5∙4 2 + 5∙4∙3 3 + 5∙4∙3∙2 4 + 5
2! 3! 4!

= 1 + 5 + 10 2 + 10 3 + 5 4 + 5

152

EXAMPLE 2
Use the binomial theorem to expand (3 + p)5.

Solution:
(3 + p)5 = [3 (1 + 3 )]5

= 35 (1 + )5

3

= 35 [1 + 5 ( ) + 5∙4 ( )2 + 5∙4∙3 ( )3 + 5∙4∙3∙2 ( )4 + ( )5]
3 2! 3 3! 3 4! 3 3

= 35 [1 + 5 ( 3 ) + 10 ( 3 )2 + 10 ( 3 )3 + 5 ( 3 )4 + ( 3 )5]

= 243 + 405 + 270 3 + 90 4 + 15 4 + 5

EXAMPLE 3
Expand (1 + − 3 2)8 up to the term in y3.

Solution:
(1 + − 3 2)8 = [1 + ( − 3 2)]8

= 1 + 8( − 3 2) + 8∙7 ( − 3 2)2 + 8∙7∙6 ( − 3 2)3 + ⋯

2! 3!

= 1 + 8( − 3 2) + 28( 2 − 6 3 + ⋯ ) + 56( 3 + ⋯ ) + ⋯

= 1 + 8 + 4 2 − 112 3 + ⋯

WORKSHEET 2.11

Q1. Expand the following using binomial theorem.

a) (2 + 1)7



Solution:

153

b) ( − √2)6
Solution:

Q2. Find the term indicated in the binomial expansions of the following functions.

a) (3 + 4 )7, 4th term. b) ( + )8 , term containing 3.

Solution: Solution:

c) (√ + 4√ )8, middle term. d) (2 3 − 1)12, term independent of x.
Solution:


Solution:

154

EXERCISE 2.5

1. Find the coefficient of 4 in the expansion of ( + 1 )5( − 1 )3.

2. Expand (1 − 3 − 2)5 in ascending powers of up to the term in 4.
2

3. Expand (2 + 3 )4 as a series of ascending powers of x.
Write down the expansion of (2 − 3 )4.
Hence, find the value of (2 + 3√2)4 − (2 − 3√2)4.

STPM PAST YEAR

STPM MT 2014 (Section A)

Q1 Use the binomial expansions of (√3 + 2)6 and (√3 − 2)6 to evaluate (√3 + 2)6 + (√3 − 2)6 .

Hence, show that 2701 < (√3 + 2)6 < 2702 . [7 marks]

STPM MM 2020 (Section A) [2 marks]
Q2 Find the general term in the expansion of (2 + 1 2)12 . [2 marks]
[4 marks]
a) Find the term independent of x in the expansion.
b) Determine the coefficient of x3 in the expansion of (1 − 3)(2 + 1 2)12 .

ANSWER 1. ⸫ The coefficient of term is -2.
Section
2.3

Exercise 1 − 15 + 35 2 − 15 3 − 515 4
2.5 2. 2 2 4 16

3. (2 + 3 )4 = 16 + 96 + 216 2 + 216 3 + 81 4
(2 − 3 )4 = 16 − 96 + 216 2 − 216 3 + 81 4

(2 + 3√2)4 − (2 − 3√2)4 = 192(√2) + 432(√2)3
= 1493.41

155

ANSWER 1 (√3 + 2)6 + (√3 − 2)6 = 2702

Section
2.3

STPM
Past Year

2
The general term, (r+1)th term = (1 2) (2 )12− ( 1 2)

(a) The term independent of x is
5 = 126720

(b) The coefficient of x3 = -14080

The Expansion of (1 + x)n , n ϵ Q .

Learning outcome: Expand (1 + x)n, where n ϵ Q, and identify the condition |x| < 1 for the validity
of this expansion.
Use binomial expansions in approximations.

(1 + ) = 1 + + ( − 1) 2 ( − 1)( − 2) 3 +⋯+ ( − 1) … ( − + 1) +⋯
2! + 3! !

The expansion is an infinite series and is called the binomial series. It is valid if |x| < 1.

The expansion is valid for (1 + x)n , not for (a + b)n.

Hence, to expand (a + b)n, it must written as ( + ) = (1 + ) .

EXAMPLE 1

Expand each of the following expressions as an ascending series in x, up to the term in x4.

State the range of x such that the expansion is valid.

1 (b) 1
(4−3 )2
(a) (1 + 2 )2

Solution:

(a) (1 + 1 = 1 + 1 (2 ) + 21(21−1) (2 )2 + 12(12−1)(21−2) (2 )3 + 21(21−1)(21−2)(12−3) (2 )4 + ⋯
2 2! 3! 4!
2 )2

= 1 + − 1 2 + 1 3 − 5 4 + ⋯

228

The expansion is valid if |2 | < 1 ,

| | < 1 or − 1 < < 1

2 22

156

(b) 1 = (4 − 3 )−2
(4−3 )2

= [4 (1 − 3 −2

4 )]

= 1 (1 − 3 −2
2 4
)

= 1 [1 + (−2) (− 3 ) + (−2)(−2−1) (− 3 2 + (−2)(−2−1)(−2−2) (− 3 3 +

2 4 2! 4 ) 3! 4 )

(−2)(−2−1)(−2−2)(−2−3) (− 3 4 + ⋯ ]

4! 4 )

= 1 [1 + 3 + 27 2 + 27 3 + 405 4 + ⋯ ]

2 2 16 16 256

= 1 + 3 + 27 2 + 27 3 + 405 4 + ⋯
2 4 32 32 512

The expansion is valid if |− 3 | <1,
4
4 4 4
| | < 3 or − 3 < < 3

EXAMPLE 2

1 in ascending power of x, where | | < 1 .

Find the first four terms in the series expansion of (1 + 2 )2 2

Hence, use the expansion to evaluate √1.02 to 5 decimal places.

Solution:

(1 + 1 = 1 + − 1 2 + 1 3 + ⋯
2 2
2 )2

Let x = 0.01, 1 = 1 + (0.01) − 1 (0.01)2 + 1 (0.01)3 + ⋯

[1 + 2(0.01)]2 22

√1.02 = 1.00995 (correct to 5 decimal places)

EXAMPLE 3

1
(1+2 )2
Expand (4−3 )2 in ascending powers of x, up to the term in x3. State the range of values of x for which

the expansion is valid.

Solution:

(1 1
(4 + 2 )2
− 3 )2 = (1 + 1 − 3 )−2

2 )2(4

⋮

= (1 + − 1 2 + 1 3 + ⋯ ) (12 + 3 + 27 2 + 27 3 + ⋯ )
2 2 4 32 32
= (1 + 3 + 27 2 + 27 3) + (1 + 3 2 + 27 3) + (− 1 2 − 3 3) + (1 3) + ⋯
24 32 32 24 32 48 4
1 5 43 25
= 2 + 4 + 32 2 + 16 3 + ⋯

The expansion is valid if |2 | < 1 and |− 3 | < 1 ,
4
| | < 1 4
and | | < 3
2
1
⸫ | | < 2

157

EXERCISE 2.6

1. In the expansion of 1+2 in ascending powers of x, the coefficient of x is zero.

1
(1+ )2

Find the value of a. With this value of a, obtain this expansion up to and including the term in x3.

2. By expressing f(x), where f(x) = 22−3 in partial fractions, or otherwise, expand f(x) as a series
6− −2 2
in ascending powers of x up to and including the term in x2.

1

3. Find the expansion of (1 − 3 )3 in ascending powers of x up to and including the term in x4.
State the range of values of x for which this expansion is valid. By taking = 18, evaluate 3√5

correct to three decimal places.

3

4. Find the expansion of (1 + 2 )2 in ascending powers of x up to and including the term in x3.

3

State the range of values of x for which this expansion is valid. By using the expansion,

evaluate (√1.004)3 correct to five decimal places.

5. Expand (1− )2 in ascending powers of y up to term in y2.

1+

1

By substituting suitable values of n and y, show that (49)8 ≈ 79601 .

51 80000

6. If is small enough for powers of higher than the third to be neglected show that

1 1 1
( + − ( − = ( + 8 33)
)2 )2 2


Use your result to deduce a rational approximation to √5 − √3 .

Hence, deduce a rational approximation to √5 + √3 .

7. Given that = 1 , where > − 1 , show that, provided ≠ 0 ,
√1+3 +√1+ 3
1
= 2 (√1 + 3 − √1 + ) .

Using this second form for y, express y as a series of ascending powers of x up to the term in x2.

Hence show, by putting = 1 , that 10 ≈ 79213 .
100 √103+√101 160000

8. Express f(x) = 2− −1 in partial fractions.
( +2)( −3)

Hence, obtain an expansion of f(x) in ascending powers of 1 up to the term in 1 .
3

Determine the set of values of x for which the expansion is valid.

158

STPM PAST YEAR

STPM MM 2014 (Section A)

Q1 Show that the first three terms in the expansions of 1 and 1−5 in ascending powers

(1 − 8 )4 1−3

of x are the same. Determine the range of values of x for which both expansions are valid.

[6 marks]

1 ≈ 1−5 1

Use the result (1 − 8 )4 1−3 to obtain an approximation of (0.84)4 as a fraction.

[3 marks]

STPM MM 2017 (Section B)

Q2 Express 20 + 5 in a single fraction. [2 marks]
1−4 1+
Hence, if x is sufficiently small such that x4 and higher powers can be neglected, show that

√ 20 + 5 ≈ 5 (1 + + 2 + 279 3 + ⋯ ) ,
1−4 1+ 16

where p and q are constants. [6 marks]

a) Determine the set of values of x for which the expansion is valid. [2 marks]

b) By substituting a suitable value of x, evaluate 6−12 correct to three significant figures.

[5 marks]

ANSWER 1 a=4
Section 1 + 2 1 = 1 + 2 2 − 8 3 + ⋯
2.3
(1 + )2
Exercise
2.6 2 ⸫ f(x) = 5 + 4
3−2 2+
5 + 4 = 11 + 1 + 67 2 + ⋯
3−2 2+ 3 9 54

3 (1 − 1 = 1 − − 2 − 5 3 − 10 4 − 22 5 + ⋯

3 )3 33 3

11
− 3 < < 3
3√5 = 1.710

4 3

(1 + 2 )2 = 1 + + 1 2 − 1 3 + ⋯
3 6 54

− 3 < < 3
2 2

1.00601 (to five decimal places)
5 (1− )2 = 1 − 4 + 8 2 2 + ⋯

1+

∴ = 1 = 1
16 50

159

6 ( + 1 − ( − 1 = 1 ( + 8 33)

)2 )2 2

129
√5 − √3 = 256

(√5 + √3) = 512
129

7 1 − 1 + 13 2 + ⋯
2 2 16

8 f(x) = 1 − 1 + 1
+2 −3
55
1 + 2 + 3 + ⋯
{ : < −3 > 3}.

ANSWER

Section 1 1

2.3 (1 − 8 )4 = 1 − 2 − 6 2 + ⋯

STPM 1 − 5 = 1 − 2 − 6 2 + ⋯
1 − 3
− 1 < < 1
Past Year
88

1 ≈ 45

(0.84)4 47

2 20 5 25
1 − 4 + 1 + = (1 − 4 )(1 + )

√ 20 + 5 = 5 (1 + 3 + 43 2 + 279 3 + ⋯ )
1−4 1+ 28 16

(a) ⸫ − 1 < < 1
4 4

(b) Let = 1 ; (6)−12 ≈ 0.331
5
(6)−12
Let = 1 ; ≈ 0.405
10
(6)−12
Let = 1 ; ≈ 0.40825
100

≈ 0.408

The answer can accurate to 5 decimal places if the substitution of x is small

enough.

160

3.1(a) MATRICES Learning Outcome:
(a) identify null, identity, diagonal, triangular and symmetric matrices;

3.1.1 MATRIX DEFINITION

Quick Notes 3.1.1 & 3.1.2

A matrix is a rectangular array of numbers enclosed between brackets.
The general form of a matrix with m rows and n columns:

11 12 13 ⋯ 1 m rows
21 22 23 ⋯ 2
31 32 33 ⋯ 3

⋮ ⋮ ⋮⋮⋮

( 1 2 3 … )

n columns

NOTE :
• The order or dimension of a matrix with m rows and n columns is m × n.
• The numbers that make up a matrix are called its entries or elements, and they are specified by

their row and column position.

• The matrix for which the entry is in ith row and jth column is denoted by ( )

EXAMPLE 1

Let = ( 5 6 1

2)
−2 3 −7

(a) What is the order of A?

(b) If A = ( ) identify 21 and 13

Solution:

(a) Since A has 2 rows and 3 columns, the order of A is 2 × 3.

(b) The entry 21 is in the second row and the first column. Thus, 21 = −2
1
The entry 13 is in the first row and the third column, and so 13 = 2

EXAMPLE 2

Given = ( )

3×3 , ≤
+ , >
Find matrix A if = {2

Solution: 31 = 2(1) + 3 = 5 123
∴ = (4 4 6)
11 = 1(1) = 1 32 = 2(2) + 3 = 7
33 = 3(3) = 9 579
12 = 1(2) = 2
13 = 1(3) = 3
21 = 2(1) + 2 = 4

161

22 = 2(2) = 4
23 = 2(3) = 6
Types of Matrices

1. Row Matrix is a (1 × n) matrix [one row]
= ( 11 12 13 … 1 )

EXAMPLE
= (2 6 7)
= (7 5 17 −1 −2)

2. Column Matrix is a (m × 1) matrix [one column]
11
21

= 31
⋮

( 1)

EXAMPLE 4
= (32) , = (3)

1

3. Square Matrix is a (n × n) matrix which has the same number of rows as
columns.

EXAMPLE

= (34 −21) , 2 × 2 matrix, 4 −3 2
= (6 1 2) , 3 × 3 matrix.
6
2 7

4. Zero Matrix is a (m × n) matrix which every entry is zero and denoted by 0.

EXAMPLE

= (00 00) 0 0 0 0 0
= (0 0) = (0 0 0)
0
0 0 0 0

5. Diagonal Matrix
The diagonal entries of A are 11, 22, ….,

11 12 13 ⋯ 1
21 22 23 ⋯ 2
Let = 31 32 33 ⋯ 3

⋮ ⋮ ⋮⋮ ⋮
( 1 2 3 … )

162

A square matrix which non-diagonal entries are all zero is called a diagonal matrix

EXAMPLE

= (10 03) 3 0 0 0 0
= (0 0 0) = (0 0)
0 0
0 2 0

6. Identity Matrix is a diagonal matrix where all its diagonal entries are 1 and denoted by I.

EXAMPLE 100
(01 01) = 2×2 (0 1 0) = 3×3

001

7. Lower Triangular Matrix is a square matrix and aij = 0 for i < j

EXAMPLE

300 0 0

= (2 −2 0) = ( 0)

142

8. Upper Triangular Matrix is a square matrix and aij = 0 for i > j

EXAMPLE

313

= (0 −2 2) = (0 )

002 0 0

Learning Outcome:
(b) use the conditions for the equality of two matrices;

3.1.2 EQUALITY OF MATRICES

Two matrices are equal if they have the same dimension, and their corresponding entries are equal.

EXAMPLE 1

Which matrices below are the same?

= (21 12) , = (1 2) , 1 2 = (21 12)
= (2 1) ,

2 1

Solution: =

EXAMPLE 2

Let = (3 − 6 42) , = (2 9 6 − 24). If A = B, find the value of a, b, c and d.
8 4 − 3 −8

163

Solution: 4 = −8 6 − = 6 2 − 3 = 8
3 − = 9 = −2 = 0 = −2

= −6

∴ = −6 , = −2 , = 0 , = −2

3.1(b) Learning Outcome:
OPERATIONS ON (c) perform scalar multiplication, addition, subtraction and multiplication of
MATRICES matrices with at most three rows and three columns;
(d) use the properties of matrix operations;

3.1.3 OPERATIONS ON MATRICES Quick Notes 3.1.3

Addition and Subtraction of Matrices
NOTE

The addition or subtraction

of two matrices with different
orders is not defined.

We say the two matrices are
incompatible.

EXAMPLE
= (31 24) B= (−45 36) C= (12)

Find:
a) + b) − c) +

Solution:

a) + = (13 42) + (−45 36)
1+4 2 + 36)
= (3 + (−5) 4 +

= (−52 150)

b) − = (31 42) − (−45 63)
1−4 2 − 63)
= (3 − (−5) 4 −

= (−83 −−12)

c) + = (13 42) + (21)

164

Since matrix A is of order 2 × 2 and matrix C is of order 2 × 1, the matrices have different orders, thus
A and C are incompatible.
Scalar Multiplication
If c is a scalar and = ( ) then = ( ), where =

EXAMPLE 1 −4 − 1
2
5 ). Find 2
Given = ( 8 7

−6

Solution:

1 1 1
− 2 = (− 2) 2 (− 2) (−4)

1 1
(− 2) 8 (− 2) 5

1 1
((− 2) (−6) (− 2) 7 )

−1 2
= −4 5
−2
(3 7
− 2)

EXAMPLE 2
Let = (15 43) and = (34 62). Calculate 3 − 2 .

Solution:
3 − 2 = 3 (15 43) − 2 (34 26)

= (135 192) − (68 142)
= (−73 05)

Properties

. + = + (Commutative)
. ( + ) + = + ( + ) (Associative)
. + (− ) = (− ) + = (0 - zero matrix)
. ( + ) = + , −
. ( + ) = +
. ( ) = ( )

165

Multiplication of Matrices

The product of two matrices A and B is defined only when the number of columns in A is equal to the

number of rows in B.
• If order of A is m × n and the order of B is n × p, then AB has order m × p.

ATTENTION!
A row and a column must have the same number of entries to be multiplied.

Multiplication Of Two Matrices 1
= ( 1 2 3 … ) 2
= 3
⋮

( )

∴ = ( 1 1 + 2 2 + 3 3 + ⋯ + )

EXAMPLE 1

Find:

(−12 2 35) 2 1
0 (−3 4)

2 1

Solution:

(−12 2 35) 2 1
0 (−3 4)

2 1 −12((11))++20((44))++35((11)))
= (−12((22))++20((−−33))++35((22))

= (62 132)

EXAMPLE 2
Let = (−31 24) and = (23 −21). Show that AB ≠ BA.

Solution:
= (−31 24) (23 −21) = (148 55)
= (23 −21) (−31 24) = (−35 104)
(148 55) ≠ (−35 104)
Thus, AB ≠ BA. This result prove that the matrix multiplication is not commutative.

166
Properties of Matrix Multiplication

Let A, B and C be matrices for which the following products are defined. Then
• Associative Property
A(BC) = (AB)C
• Distributive Property
A(B+C) = AB+AC

Transpose Matrix

Definition:

The transpose of a matrix A, written as AT, is the matrix obtained by writing each row of matrix A as s

column of A.

If × = ( ), then T × = ( ) 11 21 31
11 12 13

= ( 21 22 23) Then, = ( 12 22 32)

31 32 33 3×3 13 23 33 3×3

EXAMPLE 1 then, T = (2 1 3)1×3
2

Let = (1)

3 3×1

133 121
If = (2 5 4) then, T = (3 5 3)

135 345

EXAMPLE 2 42). Show that
Let = (13 42), = (23 41) and = (31
(a) (A + B )T = AT + BT

(b) (BC)T = CTBT

Solution: 14) = (54 65)
(a) + = (31 24) + (32
( + )T = (46 55)

T + T = (12 43) + (34 12)
= (46 55)

 (A + B)T = AT + BT

(b) = (23 14) (31 24) = (155 1200)
( )T = (2105 150)

167

T T = (14 32) (34 21) = (2105 150)
∴ ( )T = T T

Properties of Transpose Matrix
1) ( ± )T = T ± T
2) ( T)T =
3)( )T = T T
4)( )T = T ,

A Symmetric Matrix

A square matrix, = ( ) × , is symmetric if it is equal to its own transpose, A = AT that is
=

EXAMPLE 1

= (12 23) , T = (21 23)
1 − 1 −
= ( 3 ) , T = ( 3 )

− 2 − 2

A and B are symmetric matrices.

EXAMPLE 2
Given = (13 42) . Find AT. Hence, prove that AAT is a symmetric matrix.

Solution:
T = (21 43)
T = (31 42) (21 43) = (151 1251)
( T)T = (151 1215)T = (151 2115) = T

Since AAT = (AAT)T, therefore AAT is a symmetrical matrix

A Skew Symmetric Matrix

A square matrix, = ( ) × , is a skew symmetric matrix if -A = AT
= − where ≠ and = 0

168

EXAMPLE

= (20 −02) , T = (−02 20) = − (02 −02) = −

0 2 −1 0 −2 1
= (−2 0 3 ) , T = ( 2 0 −3) = −

1 −3 0 −1 3 0

A and B are skew symmetric matrix.

EXERCISE 3.1

1. Given that = (−11 23), find and such that 2 + + = where and are identity
matrix and zero matrix respectively of dimension 2 × 2.

2. If = (41 32), find
i) 2 ii) 3 iii) ( ) where ( ) = 2 3 − 4 + 3

3. Show that = (−25 −31) is a zero of polynomial ( ) = 2 − 5 + 1. Using the relationship
above, find the matrices 3 .

2 −4 3 152

4. If = ( 1 2 −1) and = (−1 −8 −5), find .

−2 −3 1 −1 −14 −8

50 0 −1 0 0

5. Matrices and are given as = (6 14 0 ), = (−2 −4 0 ).

6 3 11 −2 −1 −3
i) Determine if the matrices and abide by the communicative law of multiplication.

ii) Show that if and are numbers such that = + where is a identity matrix of

dimension 3 × 3, then find the values of and .

6. Given = (23 3 −−14) , = (10 1 −22) and = (30 0 51). Find 3 + 4 − 2 .
0 −1 −1

2 −1 and = (13 −2 −44). Find
7. Given = (1 0) 0

3 −2
i)

ii)

Is = ?

8. Find the values of , , and such that 3 ( ) = (− 1 26 ) + ( 4 + ).
+ 3


169

9. Given = (20 30) and = (70 101). Find
i) +

ii)
iii) 2 and 3 and
iv) deduce the matrix

6 −18 −4 3 2 −4

10. Given that = (1 1 2 ) and = ( 1 0 −2). Find the product of the matrices .

3 −13 −2 −2 3 3

132 32 1

11. If = (2 −1 1) and = (0 1 4 ), find .

321 1 3 −2

5 2 + 3 2

12. Matrix = ( 1 4 3 ) is symmetric. Find the values of , and .

3 − 2 6 7

1 00
13. Matrix is given by = ( 1 −2 0) for which 2 + + = . Find the values of

−1 3 1
and . Hence deduce 3.

ANSWER 1 ∴ = −4
Section ∴ = 5
3.1 2(i)
2(ii) 2 = (196 187)
Exercise 2(iii) 3 = (8441 8432)
3.1 ( ) = (18512 17567)

3 3 = (−41320 −6274)

4 300
= (0 3 0)

003

170

5(i) ∴ = , hence and abide by the communicative law of
multiplication.

5(ii) ∴ = −3
∴ = 2

6 3 + 4 − 2

= (103 13 −−1143)
−2

7(i) −1 −4 −12
= ( 1 −2 −4 )

−3 −6 −20

7(ii) = (−1182 −711)
∴ ≠

8 ∴ = 2, = 4, = 1, = 3

9(i) + = (90 104)
9(ii)
9(iii) = (104 303)

9(iv) 2 = (40 09)
10 3 = (08 207)

∴ = (20 30 )

800
= (0 8 0)

008

11 5 11 9
= ( 7 6 −4)

10 11 9

12 = 1 2 ; = 2 ; = −1

171

13 ∴ = 1 ; = −2
1 00

3 = ( 3 −8 0)
−3 9 1

3.1(c) Learning Outcome:
DETERMINANT
a) Perform the determinants of 2 X 2 matrix and 3 X 3 matrix

b) Use the concept of determinants to solve problems

Determinants of Matrices Scan for more…

1. Determinant of a 2×2 Matrices

A = ( )
The determinant of matrix A is calculated as
det A = ad – bc

EXAMPLE 1
Find the determinant of the matrix below.
A= (31 42)

Solution:
det (13 42) = (1)(4) − (2)(3)

=4–6
= -2

EXAMPLE 2
Find the determinant of the matrix below.
B = (−−52 −−34)

Solution:
det (−−52 −−34) = (−5)(−3) − (−4)(−2)

= 15 – 8
=7

172

EXAMPLE 3
Evaluate the determinant of the matrix below.
C = (−61 −32)

Solution:
det (−61 −32) = (−1)(3) − (−2)(6)

= (-3) – (-12)
= (-3) + 12
=9

EXAMPLE 4
Evaluate the determinant of the matrix below.
D = (5 − 3)

Solution:
det (5 − 3) = (5)( ) − (−3)( )

= 5 + 3

EXAMPLE 5

Find the value of in the matrix below if its determinant has a value of −12.
(−−84 2 )

Solution:

det (−−48 2 ) = −12

(−4)( ) − (2)(−8) = −12

−4 − (−16 ) = −12

−4 + 16 = −12

−4 + 16 − 16 = −12 − 16

−4 = −28

−4 −28
−4 = −4

= 7

Checking your answer: Determinant & Inverse of a matrix
Replace x by 7, then calculate the determinant. We expect to get −12.
det (−−84 27) = (−4)(7) − (2)(−8) let matrix A =[ ]

= −28 − (−16) | | = ( )( ) − ( )( )
= −
= −28 + 16
The inverse is
= −12

−

=[ − − ] [ ]

− −
= [ ] −

173

WORKSHEET 3.1

Solve for the determinant of the following 2×2 matrices.

1. (−14 −27) 2. (−−21 −63)
Solution: Solution:

3. (98 −01) 4. (−01 −111)
Solution: Solution:

174

2. Determinant of a 3×3 Matrices


A = ( )

ℎ

The determinant of matrix A is calculated as

∙ det (ℎ ) − ∙ det ( ) + ∙ det ( ℎ )
det ( )=
ℎ


EXAMPLE 1

Find the determinant of the 3×3 matrix below.
2 −3 1
(2 0 −1)
14 5

Solution:

2 −3 1

( ) = (2 0 −1)

ℎ 14 5

2 −3 1 2 ∙ det (40 −51) − (−3) ∙ det (21 −51) + 1 ∙ det (21 40)
det (2 0 −1) =
4 = 2[0 − (−4)] + 3[10 − (−1)] + 1[8 − 0]
1 5

= 2(0 + 4) + 3(10 + 1) + 1(8)

= 2(4) + 3(11) + 8

=8 + 33 + 8

= 49

EXAMPLE 2

Evaluate the determinant of the 3×3 matrix below.
132

(−3 −1 −3)
231

Solution:

132

( ) = (−3 −1 −3)

ℎ 231

1 3 2 1 ∙ det (−31 −13) − (3) ∙ det (−23 −13) + 2 ∙ det (−23 −31)
det (−3 −1 −3) =
3 = 1[−1 − (−9)] − 3[−3 − (−6)] + 2[−9 − (−2)]
2 1

= 1(−1 + 9) − 3(−3 + 6) + 2(−9 + 2)

= 1(8) − 3(3) + 2(−7)

= 8 − 9 − 14

= −15

175

EXAMPLE 3

Solve for the determinant of the 3×3 matrix below.
−5 0 −1

( 1 2 −1)
−3 4 1

Solution:

−5 0 −1

( ) = ( 1 2 −1)

ℎ −3 4 1

−5 0 −1 −5 ∙ det (24 −11) − (0) ∙ det (−13 −11) + (−1) ∙ det (−13 24)
det ( 1 2 −3) =
4 = 5[2 − (−4)] − 0[1 − (3)] − 1[4 − (−6)]
−3 1

= −5(2 + 4) − 0 − 1(4 + 6)

=−5(6) − 1(10)

=−30 − 10

= −40

WORKSHEET 3.2

Solve for the determinant of the following 3×3 matrices.
1. 1 −2 3
(−4 −5 −6)
7 −8 9

Solution:

2. −2 −1 0
( 5 1 −7)
4 −4 6

Solution:

176
3. 0 5 −6
(7 6 −3)
12 8
Solution:

4. 1 0 0
(0 1 0)
001
Solution:

5. 0 1 2
(3 0 4)
560
Solution:

177

EXERCISE 3.2 Scan for more…

A and B are 2 × 2 matrices. is the 2 × 2 multiplicative identity matrix.
1. If = , name the matrix represented by .

2. If + B = , name the matrix represented by .

3. If = , name the matrix represented by .

4. Do the matrices have inverses? Justify your answer.
a) (−−23 96)
b) (−32 69)

5. Find a value of , such that the given matrix has an inverse.

a) ( −4 39 )
2

b) (−5 5 )

Solutions:
1. = (10 01)

2. = (00 00)

3. must be the inverse matrix of .

4. a) No. The determinant (−2)(9) − (6)(−3) = 0
b) Yes. The determinant (−2)(9) − (6)(3) ≠ 0.

5. a) (−4)(9) − (3 )(2) ≠ 0
−36 − 6 ≠ 0
≠ −6.

) (5)(5) − ( )(− ) ≠ 0
25 + 2 ≠ 0
For all real values of

178

3.1.4 PROPERTIES OF Learning Outcome:
DETERMINANTS a) Properties of Determinants

Properties of Determinants

1: Switching Rows
Let A be a n × n matrix and let B be a matrix which results from switching two rows of A.
Then det(B)= − det(A).

EXAMPLE 1: Switching Two Rows
let = (13 42) and = (13 24). knowing that det (A) = - 2, find det (B)

Solution:
det(A) = 1×4−3×2 = −2. Notice that the rows of B are the rows of A but switched.
since two rows of A have been switched, det(B) = −det(A) = − (−2) = 2.

2: Multiplying a Row by a Scalar

Let A e an n × n matrix and let B be a matrix which results from multiplying some row of A by a
scalar k. Then det(B) = k det(A).

3: Adding a Multiple of a Row to Another Row
Let A be an n × n matrix and let B be a matrix which results from adding a multiple of a row to
another row. Then det(A) = det(B).

Therefore, when we add a multiple of a row to another row, the determinant of the matrix is
unchanged. Note that if a matrix A contains a row which is a multiple of another row, det(A) will
equal 0. To see this, suppose the first row of A is equal to −1 times the second row. By Theorem, we
can add the first row to the second row, and the determinant will be unchanged. However, this row
operation will result in a row of zeros.

4: Proportionality (Repetition) Property

If the all elements of a row (or column) are proportional (identical) to the elements of some other row
(or column), then the determinant is zero.

Multiplying a row by a scalar

EXAMPLE 2: Adding a Row to Another Row multiply n row of the matrix
let = (31 24) and let = (51 28). Find det (B) by k

for n=1; det(B) = k det(A)
for n=2; det(B) = k2 det(A)

for nth row;
det(B) = kn det(A)

179

Solution
det(B) = det(A) = -2.
By Definition, det(A)=−2. Notice that the second row of B is two times the first row of A added to the
second row.
⸫ det(B) = det(A)=−2.

EXAMPLE 3: Multiple of a Row
let = (21 24), show that det (A)=0

Solution
det( ) = 1 × 4 − 2 × 2 = 0.

Notice that the second row is equal to 2 times the first row. Then the determinant will equal 0.

5: Determinant of a Product

Let A and B be two n × n matrices. Then
det(AB) = det(A).det(B)

EXAMPLE 4: The Determinant of a Product

Compare det(AB) and det(A)det(B) for

= (−13 22) , = (43 21)

Solution:
= (−13 22) (34 21) = (−111 −44)
det ( ) = (−111 −44) = −40
det = det (−13 22) = 8,
det = det (34 21) = −5

det × det = 8 × −5 = −40

⸫ det ( ) = det ( ) × det( ) = 8 × −5 = −40

6: Determinant of the Transpose
Let A be a matrix where AT is the transpose of A . Then, det (AT) = det(A)

EXAMPLE 5: Determinant of the Transpose
let = (24 53), find det( )

Solution:
= (25 43)

Compute det(A) and det (AT) then compare.
det(A) = 2×3 − 4×5 = −14 and det (AT) = 2×3 − 5×4 = −14. Hence, det(A) = det (AT).

180

7: Determinant of the Inverse

Let A be an n × n matrix. Then A is invertible if and only if det(A) ≠ 0.

If this is true, it follows that

det −1 = 1
det

EXAMPLE 6: Determinant of an Invertible Matrix

Let = (32 64) , = (52 31). For each matrix, determine if it is invertible. If so, find the
determinant of the inverse.

Solution:
det(A) = 3×4 − 2×6 = 12−12 = 0

then A is not invertible.

det(B) = 2×1 − 5×3 = 2−15 = −13
Matrix B is invertible and the determinant of the inverse is given by

det( − ) = 1 Quick Notes

det( ) an inverse of a matrix
exists precisely
=1 when the determinant of
that matrix is nonzero
−13

=− 1

13

8: All-zero Property

If all the elements of a row (or column) are zero, then the determinant is zero.
9: Sum Property

1 + 1 1 1 1 1 1 1 1 1

( 2 + 2 2 2) = ( 2 2 2) + ( 2 2 2)

3 + 3 3 3 3 3 3 3 3 3

10: Triangle Property:

If all the elements of a determinant above or below the main diagonal consist of zeros, then the
determinant is equal to the product of diagonal elements. That is,

1 2 3 1 0 0

( 0 2 3) = ( 1 2 0 ) = 1 2 3
0 0 3
1 2 3

11: Reflection Property:

The determinant remains unaltered if its rows are changed into columns and the columns into rows.
This is known as the property of reflection.

181

WORKSHEET 3.3 – Determinants & Properties of Determinants

1. Explain why the matrix (64 32) has no 2. Find the inverse of (−47 −12). Show
inverse. your work. Confirm that the matrices are
inverses
Solution:
Solution:

3. Calculate the determinant of the following 4. Prove that
matrix using the properties of determinants: + 2 + 2 ) = 0
25 125 35
= ( 3 4 1 ) ( + 2
016

Solution: Solution:

5. Apply the properties of determinants and ∅
calculate the followings;
6. Show that ( ∅ ) = ( )

123 100
= (4 5 6). = (0 1 0).
Solution:
789 001

Solution:

182 8. [choose correct answer]
The determinant of resulting matrix equals to
7. [choose correct answer] k delta if elements of rows are
If determinant of a matrix A is zero then
a) added to constant k
a) A is a non-singular matrix b) divided to constant k
b) A is a singular matrix c) multiplied to constant k
c) Can’t say d) none of these
d) None of the mentioned
Solution:
Solution:

9. [choose correct answer] 10. [choose correct answer]
When two rows are interchanged, then the signs The determinant remains unaltered if its rows
of determinant are changed into columns and the columns
into rows. This is known as the property of

a) must changes b) multiplied a) reflection b) refraction
c) divided d) remain same c) repetition d) none of these

Solution:

Solution:

11. Switching property of determinant states 12a) For which values of does [ 2 + 4]
that

a) The interchange of any two rows of the have an inverse matrix
determinant changes its sign
12b) For which values of does
b) The interchange of any two columns of
the determinant changes its sign [ + 2 − 43] have an inverse matrix
− 3 +
c) The interchange of any two rows (or
two columns) of the determinant does Solution:
not changes its sign

d) Both a) and b)

Solution:

183

EXERCISE 3.3

1. Find det B given det A.


| | = |( )| = 7

ℎ


| | = |(3. 3. ℎ 3. )|



100 100

2. Given = (0 1 1) , = (0 1 0)

1 0 −2 4 001
6
and −1 = ( 2 + + ) where , ∈ ℝ, then pair of (c, d) are ______

3. Determine the determinant of the matrix A shown below:
16 2

= (3 −4 −8)

38 5
determine whether matrix A has an inverse.

4. Find the inverse matrix and verify each of following.

a) ( − )
b) ( −− )

ANSWER 1 | | = −21
Section

Exercise 2 (-6, 11)
3.3
3 det = -74
determinant of A ≠0, ⸫matrix A has an inverse

4(a) 4 3
(−151 135)
15 15
32

4(b) (47 − 57)
7 −7

184
CLONE STPM

1. Determine the values of k such that the determinant of the matrix [4 marks]

1 3
(2 + 1 −3 2) is 0.

0 2

2. The matrix A is given by
1 1

(1 2 3) and B is 3 × 1 matrix
1 1

Find the values of c for which the equation = does not have a unique solution [4 marks]

3. The matrix M and N are given by

1 1

= (1 ), = (1 )

1 3 3 3

show that det M = (a - b) (b - c) (c - a)

deduce det N. [6 marks]

ANSWER 1 = − 1 0r 2
Section 4
2
Clone 3 c = 1, c = 4
STPM

det N = -3(a-b)(b-c)(c-a)

STPM Past Year [4 marks]
[4 marks]
MM - 2017, Q3, Section A
1 −1 1

1. Matrix A is given by = (0 1 −1)
13 2

a) By using elementary row operation, determine A−1.

b) Find the determinant of A and hence, obtain the determinant of A−1

185

ANSWER 1(a) 1 10
Section 1(b) 111
3.1 −1 = − 5 5 5
1 41
STPM (− 5 − 5 5)
Pass Year

Learning Outcome:
(e) find the inverse of a non-singular matrix using elementary row
operations;

3.1.4 INVERSE MATRICES

1. Notation: Inverse matrix of = − Quick Notes 3.1.4
2. − = − =

3. Matrices with inverses are called invertible matrices.
4. Matrices with no inverse, where | | = 0, are called singular matrices.

Elementary Row Operations (EROs)

1. Interchange any two rows.

1 2 3 → 1↔ 3 7 8 9
(4 5 6) (4 5 6)

789 123

2. Multiply all the entries of a row by a scalar.

1 2 3 →2 1→ 2 2 4 6
(4 5 6) (4 5 6)

789 789

3. Multiply all the entries of a row by a scalar and add the product to

another row.

1 2 3 (→−2) 1+ 3→ 1 5 4 3
(4 5 6) (4 5 6)

789 789

Finding Inverse Matrix using Elementary Row Operations

1. To find the inverse of matrix , write the matrix in the form ( | ).
2. Change the matrix ( | ) by using elementary row operations into ( | ).
3. Matrix is the multiplicative inverse of matrix A.

186

EXAMPLE 1 3
2 −1 6), find − using elementary row operations.
1
If = ( 1 3

−2 4

Solution:

2 −1 3 1 0 0
( | ) = ( 1 3 6 | 0 1 0)

−2 4 1 0 0 1

→ 1⇌ 2 ( 1 3 6 010 [Obtain 1 in the first position on the leading diagonal]
2 −1 3 | 1 0 0)

−2 4 1 0 0 1

→ 2−2 1⟶ 2 ( 1 3 6 010
0 −7 −9 | 1 −2 0) [Obtain zeros for the rest in the first column]

−2 4 1 0 0 1

→ 3+2 1⟶ 3 1 3 60 1 0
(0 −7 −9 | 1 −2 0)

0 10 13 0 2 1

→−3 2⟶ 2 1 3 60 1 0 [Obtain 1 in the second position on the leading diagonal]
(0 21 27 | −3 6 0)

0 10 13 0 2 1

→ 2−2 3⟶ 2 1 3 6 01 0
(0 1 1 | −3 2 −2)

0 10 13 0 2 1

→ 3−10 2⟶ 3 1 3 60 1 0
(0 1 1 | −3 2 −2) [Obtain zeros for the rest in the second column]

0 0 3 30 −18 21

→ 1−3 2⟶ 1 1 0 39 −5 6
(0 1 1 | −3 2 −2)

0 0 3 30 −18 21

1 1 0 39 −5 6 [Obtain 1 in the third position on the leading diagonal]
(0 1 1 | −3 2 −2)
3→ 3⟶ 3

0 0 1 10 −6 7

→ 1−3 3⟶ 1 1 0 0 −21 13 −15
(0 1 1 | −3 2 −2 ) [Obtain zeros for the rest in the third column]

0 0 1 10 −6 7

→ 2− 3⟶ 2 1 0 0 −21 13 −15
(0 1 0 | −13 8 −9 )

0 0 1 10 −6 7

−21 13 −15
∴ − = (−13 8 −9 )

10 −6 7

187

EXAMPLE 2

−10 4 9 234

Matrices and are given as = ( 15 −4 −14) and = (4 3 1).

Find and deduce −1. −5 1 6 124

Solution:

−10 4 9 2 3 4 5 0 0
= ( 15 −4 −14) (4 3 1) = (0 5 0) = 5

−5 1 6 1 2 4 0 0 5

−1 = 5 −1 [Multiply − on the right of both sides]

= 5 −1

−1 = 1 = 1 −10 4 9 −2 0.8 1.8
5 5 ( 15 −4 −14) = ( 3 −0.8 −2.8)
1 0.2
−5 6 −1 1.2

EXERCISE 3.4

1. Find the inverse for the matrices below using elementary row operations.

−3 −1 6 −4 3 4 −3 1 2

(a) ( 2 1 −4) (b) ( 12 −9 −11) (c) ( 2 3 0)

−5 −1 11 −1 1 1 −1 1 1
012 234 2 −1 −1

(d) (2 0 1) (e) (4 3 1) (f) (−1 2 0 )

120 124 −1 0 3

2. 1 7 3 5 −7 1
If = (0 5 2 ) and = (−6 10 2 ), find AB. Hence, find − .

3 0 −1 15 −21 −5

3. 1 2 1 2 −1 −3
If = (0 1 2) and = (0 0 2 ), find AB. Hence, find − .

010 0 1 −1

4. 1 2 4
Given that = ( 4 3 8 ). Find AB if = + 4 . Hence, evaluate − .
−4 −4 −9

5. 2 3 1
Given = (−1 0 4), show that − 3 + 8 − 24 = . Deduce − .
1 −1 1

188

ANSWER 1(a) −3 −1 6 −1 −7 −5 2
Section
3.1 ∴ ( 2 1 −4) = ( 2 3 0)

Exercise −5 −1 11 −3 −2 1
3.4
1(b) −4 3 4 −1 2 13

∴ ( 12 −9 −11) = (−1 0 4)

−1 1 1 3 10

1(c) −3 1 2 −1 −3 −1 6

∴ ( 2 3 0) = ( 2 1 −4)

−1 1 1 −5 −2 11

1(d) 24 1
−9 9
0 1 2 −1 1 9 4
∴ (2 0 1) = 9 2 9
2 4 −9 2
1 0 1

( 9 9 − 9)

1(e) −2 49
3
2 3 4 −1 5 5
∴ (4 3 1) = 4 14
2 −5 −5
1 4 1 6

(−1 5 5 )

1(f) 6 3 2

2 −1 −1 −1 7 7 7
∴ (−1 2 0) = 3 5 1

−1 0 3 777
213

(7 7 7)

2 800
= (0 8 0)

008

5 7 1
−8
− = 8 5 8
1
3 4
−4 21 4
15 −8 5
− 8)
(8

189

3 200
= (0 2 0)

002

11
212
1
− = 0 2 1

1
(0 2 0)

4 −3 0 0
= ( 0 −3 0 )

0 0 −3

524
−3 −3 −3
4 7 8
− = −3 −3 −3

445

(3 3 3)

5 1 11
6 −6 2

− = 5 1 3
24 24 −8

15 1

(24 24 8 )

CLONE STPM

314
1. Given matrix = (1 1 4). Using elementary row operations, determine the inverse of A.

443

2. 1 0 0
Matrix is given by = (1 −1 0).
1 −2 1
(a) Show that 2 = , where I is the 3 × 3 identity matrix, and deduce −1.
1 43
(b) Find matrix which satisfies = ( 0 2 1).
−1 0 2

3. A, B and C are square matrices such that = − and = ( )− .
Show that − = = .
120
If = (0 −1 0), find C and A.
101

190

4. −1 2 1 −35 19 18

The matrices A and B are given by = (−3 1 4), = (−27 −13 45).

0 12 −3 12 5

Find the matrix and deduce the inverse of A.

5. The matrices P, Q and R are given by

156 −13 −50 −33 4 7 −13

= (2 −2 4) , = ( −1 −6 −5 ) , = ( 1 −5 −1 )

1 −3 2 7 20 15 −2 1 11

Find the matrices PQ and PQR, and hence, deduce ( )− .

ANSWER 1 −1 = 1 1 0
2 −2
Section 2(a) 1 7 4
3.1 2 (b) −2 26 13
4
Clone 3 (0 13 1
STPM − 13)

100
∴ −1 = (1 −1 0)

1 −2 1

8 −10 3
∴ = (3 −4 1)

1 −4 2

100
= (0 1 0)

221

1 00
= ( 0 1 0)

−2 −2 1

4 121 0 0
= ( 0 121 0 )
0 0 121

2 3 7
− 11 − 11
− = 6 11
11 2 1
− 11
3 11
(− 11 1 5
11)
11

191

5 24 40 32
= ( 4 −8 4 )

4 8 12

72 0 0
= ( 0 72 0 )

0 0 72

1 7 13
− 72
18 72
( )− = 1 5 1
− 72
72 − 72 11
1 1
72 )
(− 36 72

STPM PAST YEAR

STPM MT 2014 (Section A)
−5 0 2

1. Matrix P is given by = ( 0 2 −1).

−1 4 −2
By using elementary row operations, find the inverse of P.

STPM MT 2016 (Section A)

1 12

2. A matrix is given by = ( 0 2 2).

(a) Find 2 − 6 + 11 . −1 1 3

(b) Show that ( 2 − 6 + 11 ) = 6 , where is the 3 × 3 identity matrix and deduce −1.

ANSWER

Section 1. 0 2 −1
∴ −1 = (0.25 3 −1.25)
3.1
0.5 5 −2.5

STPM 4 −1 −2
Past Year 2(a) 2 − 6 + 11 = (−2 5 −2)

2(b) 2 −2 2

2 11
3 −6 −3

−1 = 1 5 1
−3 6 −3

1 11
( 3 −3 3 )

192 Learning Outcomes:
a) Reduce an augmented matrix to row-echelon form, and determine
3.2 SYSTEMS OF whether a system of linear equations has a unique solution, infinitely
LINEAR many solutions or no solution.
EQUATIONS b) Find the unique solution of a system of linear equations using the
inverse of a matrix.

Unique, Inconsistent and Infinite Solutions Quick Notes 3.4
In two- dimensional geometry, when 2 lines are given, they must be
intersecting, or parallel, or coincident.

Intersecting lines
2x+4y=5
4x-3y=2

There is one point of intersection and the system is said to be consistent
and has a unique solution.

Parallel lines
3x+y=2
3x+y=6

There is no point of intersection and in this case there is no solution
and the system is said to be inconsistent.

Coincident lines
x+3y=1

3x+9y=3
There are infinitely many points of intersection, thus the system is said
to be consistent and has an infinite number of solutions.

System of Linear Equations in Matrix Notation
The set of equations below is called a system of linear equations in three
variables, x, y and z.

11 + 12 + 13 = 1
21 + 22 + 23 = 2
31 + 32 + 33 = 3

11 12 13 1

If A = ( 21 22 23) , X = ( ) and B = ( 2) , then the given equations can be written as = .
31 32 33
3

For example, the equations

3x-2y+4z = 4 2x+3y-2z = 2 4x+2y-3z = 3

can be written in matrix notation as AX=B where
x
3 −2 4 4
A = (2 3 −2) , X = (y) and B = (2)
z
4 2 −3 3

193

Finding the unique solution using the inverse matrix

Consider a system of n linear equations with n variables whose matrix of coefficients A has an inverse

A-1.

AX = B (given system)

A-1(AX) = A-1B (pre-multiply both sides by )

(A-1A)X = A-1B (associative property)

IX = A-1B (inverse property)

X = A-1B (identity property)

Therefore, to find X, simply multiply A-1 and B.

Solving system of Linear Equations (Gaussian Elimination Method)

1. Express system of linear equations into augmented matrix (AM) form.
2. Using Elementary Row Operations (ERO), reduced the left (AM) into Row Echelon Form (REF)

a) entries (2,1) and (3,1) are zeroes, THEN
b) entry (3,2) is zero.

3. Rewrite the reduced AM into system of linear equations and solve them in reverse order

(Solve Row 3 → Solve Row 2 → Solve Row 1)

Gaussian elimination Method

1. If we were to have the following system of linear equations containing three equations for three

unknowns:

x+ y+ z=3

x + 2y + 3z = 0 → {system of linear equations to solve

x + 3y + 2z = 3

2. We know from our lesson on representing a linear system as a matrix that we can represent such
system as an augmented matrix like below:

x+ y+ z=3 → 1 1 1 3 {Transcribing the linear system into an augmented matrix
x + 2y + 3z = 0 1 2 3 0

x + 3y + 2z = 3 1 3 2 3

3. Let use Gaussian elimination Method so we can simplify the matrix

⎯⎯⎯⎯→1 1 1 3 1 1 1 3 
0 1 2 − 3
1 2 3 0
R2 −R1 →R2

1 3 2 3  1 3 2 3 
 

⎯⎯⎯⎯→R3 −R1→R3 1 1 1 3 
0 1 2 − 3
 
 0 2 1 0 

194

⎯⎯⎯⎯→2R2 →R2 1 1 1 3 
0 2 4 − 6
 
 0 2 1 0 

⎯⎯⎯⎯→R3 −R2 →R3 1 1 1 3 
0 2 4 − 6 {Reduced matrix into its echelon form
 
 0 0 −3 6 

4. Resulting linear system of equations to solve
x+ y+z=3

2 y + 4z = −6

− 3z = 6

5. (Solve Row 3 → Solve Row 2 → Solve Row 1) {final solution.
Solve Row 3 → z = −2
Solve row 2 → 2y + 4(− 2) = −6  y =1
Solve row 1 → x + 1 + (− 2) = 3  x = 4

Difference between Gaussian Elimination and Gaussian Jordan Elimination

1. Gaussian Jordan Elimination Method is to reduce the left Augmented Matrix (AM) to Reduced
Row Echelon Form (RREF).

2. Row Echelon Form (REF) →  a b c 
0 d e

0 0 f 

3. Reduced Row Echelon Form (RREF) →  1 0 0 
0 1 0

0 0 1

EXAMPLE 1: Solving Systems Linear Equations of By Elementary Row Operations

Solve the following system of linear equations.

x + 2y + 7z = 1 x + 3y = 2 -y + 8z =3

Solution:

Writing in matrix notation
x
127 1
A = (1 3 0) , X = (y) and B = (2)
z
0 −1 8 3

To find inverse of matrix A using elementary row operation.

195

1 2 71 0 0 1 2 71 00
(1 3 0|0 1 0) (−1) 1 + 2 → 2 (0 1 −7|−1 1 0)

0 −1 8 0 0 1 0 −1 8 0 0 1

(−2) 2 + 1 → 1 1 0 21 3 −2 0
2 + 3 → 3 (0 1 −7|−1 1 0)

0 0 1 −1 1 1

(−21) 3 + 1 → 1 1 0 0 24 −23 −21
7 3 + 2 → 2 (0 1 0|−8 8 7 )

0 0 1 −1 1 1

24 −23 −21
A−1 = (−8 8
7)

−1 1 1

Therefore, X = A-1B
24 −23 −21

= (−8 8 7 ) ( )

−1 1 1

−85
= ( 29 )

4

Hence, x = -85, y = 29 and z = 4.

EXAMPLE 2: Finding Inverse By Elementary Row Operations

1 −2 1
If P = ( 3 1 2) , find P-1.

−1 4 1

Hence, solve the simultaneous equations
x – 2y + z =1

3x + y + 2z = 4

-x + 4y + z = 2

Solution: 1 −2 1 1 0 0
The augmented matrix is (0 7 −1|−3 1 0)

1 −2 1 1 0 0 0 2 21 01
( | ) = ( 3 1 2|0 1 0)

−1 4 1 0 0 1

2 − 3 1 → 2

3 + 1 → 3