Semester 1 Mathematics STPM (Teacher's Edition)

196

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 ) + 12(12−1) (2 )2 + 12(12−1)(12−2) (2 )3 + 21(21−1)(21−2)(21−3) (2 )4 + ⋯

2 )2 2 2! 3! 4!

= 1 + − 1 2 + 1 3 − 5 4 + ⋯
2 2 8

The expansion is valid if |2 | < 1 ,

| | < 1 or − 1 < < 1

2 22

(b) 1 = (4 − 3 )−2
(4−3 )2

= [4 (1 − 3 −2
4
)]

= 1 (1 − 3 −2

24 )

= 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 + ⋯
24 32 32 512

The expansion is valid if |− 3 | <1,

4
| | < 4 − 4 < < 4
or
3 33

197

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 + ⋯
2 2
[1 + 2(0.01)]2

√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 + ⋯ ) (1 + 3 + 27 2 + 27 3 + ⋯ )
22 24 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 2 + 25 3 + ⋯
24 32 16
3
The expansion is valid if |2 | < 1 and |− 4 | < 1 ,

| | < 1 and | | < 4
2 3
⸫ | | < 1
2

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 = 1, evaluate 3√5

8

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.

198

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.

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
1−3
(1 − 8 )4

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 6−21 correct to three significant figures.

[5 marks]

199 1 1+2 = (1 + 2 )(1 + )−12

ANSWER 1
Section (1+ )2
2.3
= (1 + 2 ) [1 + (− 1) ( ) + (−12)(−32) ( )2 + (−12)(−32)(−52) ( )3 +
Exercise 2 2! 3!
2.6
⋯]

= (1 + 2 ) (1 − + 3 2 2 − 5 3 3 + ⋯ )
28 16
= (1 − + 3 2 2 − 5 3 3 + ⋯ ) + (2 − 2 + 3 2 3 + ⋯ )
28 16 4
= 1 + (2 − ) + (3 2 − ) 2 + (3 2 − 5 3) 3 + ⋯
28 4 16
Since the coefficient of x is zero, therefore 2 − = 0
2

a=4

When a = 4, 1+2 = 1 + (3(4)2 − 4) 2 + (3(4)2 − 5(4)3) 3 + ⋯

1 8 4 16

(1+ )2

= 1 + 2 2 − 8 3 + ⋯

2 22−3 = 22−3
6− −2 2 (3−2 )(2+ )

Let 22−3 = +
(3−2 )(2+ )
3−2 2+

22 − 3 = (2 + ) + (3 − 2 )

Substituting x = -2; 28 = B(7)

B=4

Substituting x = 3 ; 22 − 9 = (2 + 32)
2 2
35 7
2 = 2

A=5

⸫ f(x) = 5 + 4
3−2 2+
5 + 4 = 5(3 − 2 )−1 + 4(2 + )−1
3−2 2+
−1 2 )−1
= 5 (1 − 2 + 2 (1 +
3 3 )

= 5 [1 + (−1) (− 2 ) + (−1)(−2) (− 2 2 + ⋯ ] + 2 [1 +
3 3 2! 3
)

(−1) ( 2 ) + (−1)(−2) ( 2 )2 + ⋯ ]
2!

= 5 (1 + 2 + 4 2 + ⋯ ) + 2 (1 − 1 + 1 2 + ⋯ )
3 39 24
11 1 67
= 3 + 9 + 54 2 + ⋯

This question can change to,

…expand f(x) as a series in ascending powers of 1 up to and including the


1
term in 3 .

⸫ f(x) = 5 + 4
3−2 2+
5 + 4 = 5(3 − 2 )−1 + 4(2 + )−1
3−2 2+
= 5(−2 + 3)−1 + 4( + 2)−1

= 5 [(−2 ) (1 − 23 )]−1 + 4 [ (1 + 2 )]−1
−1 2)−1
= (− 5 ) (1 − 3 + 4 (1 +
)
2 2

200

= (− 5) [1 + (−1) (− 3) + (−1)(−2) (− 3 2 + ⋯ ] +
2!
2 2 2 )

4 [1 + (−1) ( 2 ) + (−1)(−2) (− 2 )2 + ⋯ ]
2!

= (− 5) (1 + 3 + 9 + ⋯ ) + 4 (1 − 2 + 4 + ⋯ )
2 4 2 2
2 15 16
5 4 2 45 ( 4 8 3
= (− 2 − − 8 3 + ⋯ ) + − 2 + + ⋯ )

= 3 ( 1 ) − 47 ( 1 2) + 83 ( 1 3) + ⋯
2 4 8

3 (1 − 1 = 1 + 1 (−3 ) + + (31)(−23) (−3 )2 + (13)(−23)(−35) (−3 )3 +

3 )3 3 2! 3!

(13)(−23)(−53)(−83) (−3 )4 + (13)(−23)(−53)(−83)(−131) (−3 )5 + ⋯
4! 5!
= 1 − − 2 − 5 3 − 10 4 − 22 5 + ⋯
33 3

The expansion is valid for |−3 | < 1

1
| | < 3
11
− 3 < < 3
1
Substituting = 1 ; = 1 − (1) − (1)2 − 5 (1)3 − 10 (1)4 −
8 [1 − 3 (1)]3
8 8 8 38 38
22 (1)5 + ⋯
38
1

(5)3 = 1 − 1 − 1 − 5 − 5 − 11 + ⋯

8 8 64 1536 6144 49152

3√5 = 0.855082

2

3√5 = 1.710

4 (1 + 3 = 1 + 3 (2 ) + (23)(21) (2 )2 + (23)(21)(−21) (2 )3 + ⋯

2 )2 3! 3
3 23 2! 3
= 1 + + 1 2 − 1 3 + ⋯
6 54
2
The expansion is valid for | 3 | < 1

| | < 3
2
− 3 < < 3
22
(√1.004)3 3
=
(1.004)2

3

= (1 + 0.004)2

2 = 0.004
3
⸫ x = 0.006

= 1 + (0.006) + 1 (0.006)2 − 1 (0.006)3 + ⋯
6 54

= 1.00601 (to five decimal places)

5 (1− )2 = (1 − )2 (1 + )−2

1+

= [1 + (2 )(− ) + (2 )(2 −1) (− )2 + ⋯ ]
2!
(2 )(2 −1)
[1 + (2 )(− ) + 2! (− )2 + ⋯ ]

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

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

= 1 − 4 + 8 2 2 + ⋯

Let 1− = 49
51
1+

201

51 − 51 = 49 + 49

2 = 100
= 1

50

∴ = 1 = 1

16 50

(1− )2 ≈ 1 − 4 + 8 2 2

1+

(11−+551100)2(116) ≈ 1 − 4 (116) (510) + 8 (116)2 (510)2
1
(5419⁄⁄5500)8
≈ 1 − 1 + 1
200 80000

1
(5419⁄⁄5500)8
≈ 79601
80000

6 ( + 1 − ( − 1 = 1 (1 + 1 − 1 (1 − 1

)2 )2 2 )2 2 )2

(21)(−21) (21)(−21)(−23)
= 1 [1 + (1) ( ) + ( )2 + ( )3 + ⋯ ] −
2 2! 3!
2

1 [1 + (1) (− ) + (21)(−21) (− 2 +

2 2 2! )

(12)(−12)(−32) (− )3 + ⋯ ]

3!

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

2 2 8 16 ) 2 2

1 ( 2 − 1 ( )3 + ⋯ ]

8 ) 16

= 1 [1 ( ) + 1 ( ) + 1 ( 3 + 1 ( )3]

2 2 2 16 ) 16

= 1 ( + 8 33)

2

Let a = 4, b = 1 ; ( + 1 − ( − 1 = 1 ( + 8 33)

)2 )2 2

(4 + 1 − (4 − 1 = 1 (1 + 8((14))33)

1)2 1)2 42 4

√5 − √3 = 2 (1 + 1)

4 512
= 129
256

(√5 − √3)(√5 + √3) = 129 (√5 + √3)
256
2 (√3)2
− = 129 (√5 + √3)
(√5) 256

2 (256) = (√5 + √3)

129 512
129
(√5 + √3) =

7 = 1 × √1+3 −√1+
√1+3 +√1+ √1+3 −√1+

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

202

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

= 1 {[1 + 1 (3 ) + (12)(−12) (3 )2 + (12)(−21)(−23) (3 )3 + ⋯ ] − [1 + 1 ( ) +
2 2 2! 3! 2

(12)(−12) ( )2 + (21)(−12)(−23) ( )3 + ⋯ ]}

2! 3!

= 1 {[1 + 3 − 9 2 + 27 3 + ⋯ ] − [1 + 1 − 1 2 + 1 3 + … ]}
2 28 16 28 16
1 13
= 2 ( − 2 + 8 3 + ⋯ )

= 1 − 1 + 13 2 + ⋯
2 2 16

= 1 = 1 (√1 + 3 − √1 + )
√1+3 +√1+ 2
∴ 1 = 1 − 1 + 13 2 + ⋯
√1+3 +√1+ 2 2 16

Let = 1 ; 1 ≈ 1−1( 1 ) + 13 ( 1 2
100
√1+3(1010)+√1+(1010) 2 2 100 16 100 )

1 ≈ 1 − 1 + 13

√110030+√110001 2 200 160000

10 ≈ 79213
√103+√101 160000

8 f(x) = 2− −1
( +2)( −3)

= 2− −6+5
2− −6
5
= 1 + ( +2)( −3)

Let 5 = +
( +2)( −3) +2 −3

5 = ( − 3) + ( + 2)

Substituting x = -2 ; 5 = A(-5)

A = -1

Substituting x = 3 ; 5 = (5)

B=1

Hence, f(x) = 1 − 1 + 1
+2 −3
1 1
= 1 − (1+ 2 ) + (1− 3 )

= 1 + 1 [− (1 + 2)−1 + (1 − 3)−1]


= 1 + 1 {− [1 + (−1) 2 + (−1)(−2) ( 2 )2 + ⋯ ] + [1 +
2!

(−1) (− 3) + (−1)(−2) (− 3)2 + ⋯ ]}

2!

= 1 + 1 [− (1 − 2 + 4 + ⋯ ) + (1 + 3 + 9 + ⋯ )]
2 2
1 (5 5
= 1 + + 2 + ⋯ )

5 5
= 1 + 2 + 3 + ⋯

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


| | > 2 and | | > 3

⸫ | | > 3

The set of values of x for which the expansion to be valid is { : <

−3 > 3}.

203

ANSWER

Section 1 1 1 (14) (− 34)
2.3 ()
(1 − 8 )4 = 1 + (−8 ) + (−8 )2 + ⋯
4 2!
= 1 − 2 − 6 2 + ⋯
STPM

Past Year 1 − 5 = (1 − 5 )(1 − 3 )−1
1 − 3
(−1)(−2)
= (1 − 5 ) [1 + (−1)(−3 ) + 2! (−3 )2 + ⋯ ]

= (1 − 5 )(1 + 3 + 9 2 + ⋯ )

= (1 + 3 + 9 2 + ⋯ ) + (−5 − 15 2 + ⋯ )

= 1 − 2 − 6 2 + ⋯

Hence, the first 3 terms in the two expansions in ascending powers of x are

the same.

Both expansions are valid if |−8 | < 1 and |−3 | < 1 ,

| | < 1 and | | < 1

8 3
− 1 < < 1 − 1 < < 1
and
88 33
1 1
⸫ − 8 < < 8

1 − 8 = 0.84

x = 0.02

Given (1 − 1 ≈ 1−5

8 )4 1−3

[1 − 1 ≈ 1−5(0.02)
1−3(0.02)
8(0.02)]4

1 0.9

(0.84)4 ≈ 0.94

≈ 90
94
≈ 45
47

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

25
= (1 − 4 )(1 + )

√ 20 + 5 = √(1−4 2 )5(1+ )
1+
1−4

= 5

11
(1−4 )2(1+ )2

= 5(1 − 4 )−12(1 + )−12

[1 + (− 1) (−4 ) + (−21)(−32) (−4 )2 + (−21)(−23)(−25) (−4 )3 + ⋯ ]
2 2! 3!
=5
{ [1 + (− 1) ( ) + (−21)(−32) ( )2 + (−21)(−23)(−52) ( )3 + ⋯ ]
2 2! 3! }

= 5 {(1 + 2 + 6 2 + 20 3 + ⋯ ) (1 − 1 + 3 2 − 5 3 + ⋯ )}
28 16
= 5 (1 − 1 + 3 2 − 5 3 + 2 − 2 + 3 3 + 6 2 − 3 3 + 20 3 + ⋯ )
28 16 4
= 5 (1 + 3 + 43 2 + 279 3 + ⋯ )
28 16

204

(a) The expansion is valid if |−4 | < 1 and | | < 1 ,

| | < 1 and | | < 1

4
− 1 < <1
and −1 < < 1
4 4
1 1
⸫ − 4 < < 4

The set of values of x for which the expansion to be valid is { : − 1 < <
4
1
4 } .

(b) Let = 1 ; 5 (1 − 4)−21 (1 + 1)−21 = 5 [1 + 3 (1) + 43 (1)2 + 279 (1)3 + ⋯ ]
5
55 25 85 16 5

25(6)−21 ≈ 5 (3309)

2000

(6)−12 ≈ 1 (3309)

5 2000

≈ 0.331

OR

Let = 1 ; 5 (1 − 4 )−12 (1 + 1 )−21 = 5 [1 + 3 ( 1 ) + 43 ( 1 2 + 279 ( 1 3 +
10
10 10 2 10 8 10 ) 16 10 )

⋯]

5 ( 6 )−21 (11)−12 ≈ 5 (19539)
10 10 16000
1
(6)−21 ≈ (19539)
(11)2

10 16000

≈ 0.405

OR

Let = 1 ; 5 (1 − 4 )−12 (1 + 1 )−12 = 5 [1 + 3 ( 1 ) + 43 ( 1 2
100
100 100 2 100 8 100 )+

279 ( 1 3 + ⋯ ]
16
100 )

5 ( 96 )−21 (101)−21 ≈ 5(1.01555)
100 100
(16×6)−12 (101)−21 ≈ (1.01555)
100 100
1
(6)−21 ≈ (1.01555)
4(101)2

100

≈ 0.40825

≈ 0.408
Note: From calculator (6)−21 = 0.40825

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

enough.

205

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
22 = 2(2) = 4
23 = 2(3) = 6

206

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
= (23) , = (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 … )

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

EXAMPLE

= (10 30) 3 0 0 0 0
= (0 0 0) = (0 0)
0 0
0 2 0

207

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

EXAMPLE 100
(01 10) = 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?

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

2 1

Solution: =

EXAMPLE 2

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

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

= −6

∴ = −6 , = −2 , = 0 , = −2

208

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
= (13 24) B= (−45 36) C= (12)

Find:
a) + b) − c) +

Solution:

a) + = (31 24) + (−45 63)
1+4 2 + 63)
= (3 + (−5) 4 +

= (−52 150)

b) − = (31 24) − (−45 63)
1−4 2 − 63)
= (3 − (−5) 4 −

= (−83 −−12)

c) + = (31 42) + (12)

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 =

209

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

−6

Solution:

1 (− 1 (−4)
(− 2) 2 2)
1
− 2 = 1 1
(− 2) 8 (− 2) 5

((− 1 (−6) 1
2) (− 2) 7 )

−1 2
= −4 5
−2
(3 7
− 2)

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

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

= (135 192) − (68 142)
= (−73 50)

Properties

. + = + (Commutative)
. ( + ) + = + ( + ) (Associative)
. + (− ) = (− ) + = (0 - zero matrix)
. ( + ) = + , −
. ( + ) = +
Multip l.ica t(io n o) f=M(a t r i)c e s

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.

210

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

( )

∴ = ( 1 1 + 2 2 + 3 3 + ⋯ + )

EXAMPLE 1

Find:

(−12 2 53) 2 1
0 (−3 4)

2 1

Solution:

(−12 2 35) 2 1
0 (−3 4)

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

= (26 132)

EXAMPLE 2
Let = (−31 24) and = (32 −21). Show that AB ≠ BA.

Solution:
= (−31 42) (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.

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

211

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), = (32 14) and = (13
(a) (A + B )T = AT + BT

(b) (BC)T = CTBT

Solution: 41) = (54 65)
(a) + = (13 24) + (32
( + )T = (64 55)

T + T = (21 34) + (43 21)
= (64 55)

 (A + B)T = AT + BT

(b) = (32 41) (13 42) = (155 1200)
( )T = (2150 150)
T T = (14 32) (34 21) = (2150 150)

∴ ( )T = T T

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

212

A Symmetric Matrix

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

EXAMPLE 1

= (12 32) , T = (21 32) 1
1 − −
= ( 3 ) , T = ( 3 )

− 2 − 2

A and B are symmetric matrices.

EXAMPLE 2
Given = (31 24) . Find AT. Hence, prove that AAT is a symmetric matrix.

Solution:
T = (21 43)
T = (31 24) (21 43) = (151 2151)
( T)T = (151 1251)T = (151 1251) = 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

EXAMPLE

= (20 −02) , T = (−02 02) = − (20 −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.

213

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 = (14 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 −−41) , = (01 1 −22) and = (30 0 51). Find 3 + 4 − 2 .
0 −1 −1

2 −1 and = (31 −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


9. Given = (02 03) and = (07 101). 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

214

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 2 + + =
Section (−11 32) (−11 23) + (−11 32) + (10 01) = (00 00)
3.1 (−−14 87) + (− 32 ) + ( 0 0 ) = (00 00)
−4 − = 0 , ∴ = −4
Exercise
3.1 −1 + + = 0 , ∴ = 5

2(i) 2 = (41 32) (14 32) = (196 187)
2(ii)
2(iii) 3 = 2 = (196 187) (14 32) = (8414 8432)

( ) = 2 3 − 4 + 3 23) + (03 03)

= 2 (4841 8432) − 4 (41
= (18512 17567)

3 ( ) = 2 − 5 + 1

( ) = 2 − 5 +

= (−25 −31) (−25 −31) − 5 (−25 −31) + (01 01)
= (−925 −145) − (−1205 −155) + (01 10)
= (00 00)
Therefore = (−25 −31) is a zero of polynomial

215

( ) = 2 − 5 + 1.

2 − 5 + = 0

Multiplying both sides by

2 − 5 + = 0

3 − 5 2 + = 0

3 = 5 2 −

= 5 (−925 −145) − (−25 −31)
= (−41320 −6274)

4 2 −4 3 1 5 2
= ( 1 2 −1) (−1 −8 −5)
−2 −3 1 −1 −14 −8

300
= (0 3 0)

003

5(i) 5 0 0 −1 0 0
= (6 14 0 ) (−2 −4 0 )
6 3 11 −2 −1 −3

−5 0 0
= (−34 −56 0 )

−34 −23 −33

−1 0 0 5 0 0
= (−2 −4 0 ) (6 14 0 ) =

−2 −1 −3 6 3 11

−5 0 0
(−34 −56 0 )

−34 −23 −33

∴ = , hence and abide by the communicative law of
multiplication.

5(ii) = +

50 0 −1 0 0 100

(6 14 0 ) = (−2 −4 0 ) + (0 1 0)

6 3 11 −2 −1 −3 001

6 = −2 , ∴ = −3

5 = − + , ∴ = 2

216

6 3 + 4 − 2

= 3 (32 3 −−14) + 4 (10 1 −22) − 2 (03 0 15)
0 −1 −1

= (96 9 −−132) + (40 4 −88) − (60 0 120)
0 −4 −2

= (103 13 −−1143)
−2

7(i) 2 −1 (31 −2 −44) −1 −4 −12
= (1 0) 0 (1 −2 −4 )
= −6
3 −2 −3 −20

7(ii) = (31 −2 −44) 2 −1 = (−1182 −711)
0 (1 0)

∴ ≠ 3 −2

8 3 ( ) = (− 1 26 ) + ( 4 + )
+ 3


3 = + 4, ∴ = 2

3 = 6 + + ; 2 = 8 , ∴ = 4

3 = 2 + 3 , ∴ = 3

3 = −1 + + ; 2 = 2 , ∴ = 1

∴ = 2, = 4, = 1, = 3

9(i) + = (02 03) + (70 101) = (90 104)
9(ii)
9(iii) = (20 30) (07 101) = (104 303)

9(iv) 2 = (02 03) (02 03) = (40 09) 207)
10 3 = 2 = (04 90) (02 03) = (80

2 = (202 0 ) ; 3 = (203 303) ; ∴ = (20 0 )
32 3

6 −18 −4 3 2 −4 8 0 0

= (1 1 2 ) ( 1 0 −2) = (0 8 0)

3 −13 −2 −2 3 3 008

11 1 3 2 32 1 5 11 9

= (2 −1 1) (0 1 4 ) = ( 7 6 −4)

3 2 1 1 3 −2 10 11 9

217

12 5 2 + 3 2

= ( 1 4 3 )

3 − 2 6 7

By the property of symmetry,

2 + 3 = 1 ; ∴ = −1

3 = 6 ; ∴ = 2

2 = 3 − 2

2 − 3 + 2 = 0

( − 1)( − 2) = 0

∴ = 1 2

= 1 2 ; = 2 ; = −1

13 2 + + =

1 00 1 00 1 00

( 1 −2 0) ( 1 −2 0) + ( 1 −2 0)

−1 3 1 −1 3 1 −1 3 1

100 000

+ (0 1 0) = (0 0 0)

001 000

1 00 0 0 0 0

(−1 4 0) + ( −2 0 ) + (0 0)

1 −3 1 − 3 0 0

100

= (0 1 0)

001
−1 + = 0 ; = 1

1 + + = 0 ; = −2

∴ = 1 ; = −2

2 + − 2 =

Multiplying both sides by

2 + − 2 =

3 + 2 − 2 = 0

3 = − 2 + 2

1 00 1 00

= − (−1 4 0) + 2 ( 1 −2 0)

1 −3 1 −1 3 1
1 00

= ( 3 −8 0)

−3 9 1

218 Learning Outcome:

3.1(c) a) Perform the determinants of 2 X 2 matrix and 3 X 3 matrix
DETERMINANT
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 (31 24) = (1)(4) − (2)(3)

=4–6
= -2

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

Solution:
det (−−52 −−43) = (−5)(−3) − (−4)(−2)

= 15 – 8
=7

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

219

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:

Replace x by 7, then calculate the determinant. We expect to get −12.

det (−−48 72) = (−4)(7) − (2)(−8)
= −28 − (−16)
Determinant & Inverse of a matrix

= −28 + 16

= −12 let matrix A =[ ]

| | = ( )( ) − ( )( )
= −

The inverse is −


=[ − − ] [ ]

− −
= [ ] −

220

WORKSHEET 3.1

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

1. (−14 −27) 2. (−−12 −63)

Solution: Solution:

det (−14 −27) = (1)(−7) − (2)(−4) det (−−12 −63) = (−1)(6) − (−3)(−2)
= −7 − (−8) = −6 − (6)
= −7 + 8 = −12
=1

Ans : 1 Ans : -12

3. (89 −01) 4. (−01 −111)

Solution: Solution:

det (89 −01) = (8)(−1) − (0)(9) det (−01 −111) = (0)(−11) − (1)(−1)
= −8 − 0 = 0 − (−1)
= −8 =0+1
=1

Ans : -8 Ans : 1

2. Determinant of a 3×3 Matrices


A = ( )

ℎ

The determinant of matrix A is calculated as

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


221

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 04)
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

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

222

−5 0 −1 −5 ∙ det (24 −11) − (0) ∙ det (−13 −11) + (−1) ∙ det (−13 42)
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: 3 = 1 ∙ det (−−85 −96) − (−2) ∙ det (−74 −96) +
1 −2 −6)
3 ∙ det (−74 −−58)
det(−4 −5 9

7 −8

= 1[(−45) − 48] + 2[−36 − (−42)] + 3[32 − (−35)]

= 1(−93) + 2(6) + 3(67)

= −93 + 12 + 201

= 120

2. −2 −1 0
( 5 1 −7)

4 −4 6

Solution: 0 = −2 ∙ det (−14 −67) − (−1) ∙ det (45 −67) +
−2 −1 −7)
0 ∙ det (45 −14)
det ( 5 1 6

4 −4

= −2[6 − 28] + 1[30 − (−28)] + 0

= −2(−22) + 1(58) + 0

= 44 + 58 + 0

= 102

3. 0 5 −6
(7 6 −3)

12 8

Solution: −6 = 0 ∙ det (62 −83) − 5 ∙ det (71 −83) + (−6) ∙ det (71 62)
05 −3)

det(7 6 8

12

223

= 0−5[56 − (−3)] + (−6)[14 − 6)]
= 0−5(59) + (−6)(8)
= −343

4. 1 0 0
(0 1 0)

001
Solution:

100
det (0 1 0) = 1. (1 − 0) − 0(0 − 0) + 0(0 − 0)

001
=1

5. 0 1 2
(3 0 4)

560

Solution: −2
0 −1 4 ) = 0(0 − 24) + 1(0 − 20) − 2(18 − 0)
0
det (3 0
= 0 – 20 – 36
56
= -56

EXERCISE 3.2

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 69)
b) (−32 69)

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

a) ( −4 39 )
2

b) (−5 5 )

224 Scan for more…

Solutions:
1. = (01 10)

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

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 = (31 42) and = (31 42). 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). Multiplying a row by a scalar

multiply n row of the matrix
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)

225

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.

EXAMPLE 2: Adding a Row to Another Row
let = (31 24) and let = (51 28). Find det (B)

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 12)

Solution:
= (−13 22) (43 12) = (−111 −44)
det ( ) = (−111 −44) = −40

226

det = det (−13 22) = 8,
det = det (43 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).

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 46) , = (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

det( )

=1

−13

=− 1

13

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

227

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.

Quick Notes

an inverse of a matrix
exists precisely
when the determinant of
that matrix is nonzero

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:
Determinant (46 32) ( ) (−47 −12) = (10 10)
= (6)(2) − (3)(4) = 0

This means the area of the image is 0 because −7 + 4 = 1, −2 +
the image of the unit square maps to a straight = 0, −7 + 4
= 0, −2 + = 1
line, which has no area. This also means that
= 1, = −4, = 2, and = −7
distinct points map to the same location, so the (−47 −12)−1 = (−14 −27)

transformation cannot be reversed. so, the matrix
[64 23] has no inverse.

(−14 −27) (−47 −12) = (10 01)

228

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(5) 5(25) 5(7)
| | = | 3 4 1 |. ( + 2 + 2 0 + 2 ).

016

5 25 7
| | = 5 |3 4 1| //
= ( ) + (2 2 2 )
016
[scalar multiple property]

| | = 5(−314) = −1570.

= ( ). //


[proportionality property]

=0

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: ∅

. . = ( ∅ ) = ( )

123

| | = |4 5 6| (Interchanging rows and columns across the

789 diagonal) //reflection property

//[proportionality/ ∅

repetition property] = (−1) ( ∅ ) = (−1)2 ( )

r2 > r1 + 3

r3 > r2 + 3 //switching property

| | = 0 ∅

= ( ) = . .



100
= (0 1 0) //Triangular property

001
| | = 1(1 − 0) + 0(0 − 0) + 0(0 − 0)
| | = 1

229

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

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:
a) a)

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 12a)

d) Both a) and b) ( )( ) − ( + )( ) ≠
− − ≠
Solution: ( − )( + ) ≠
Both a) and b) ≠ , ≠ −

12b)

( + 2)( + 3) − ( − 4)( − 3) ≠ 0
( 2 + 5 + 6) − ( 2 − 7 + 12) ≠ 0

12 − 6 ≠ 0
1

≠ 2

230

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
Section |A| = |( )| = 7

Exercise ℎ
3.3

|B| = |(3. 3. ℎ 3. )|





|B| = |(3. 3. ℎ 3. )| = (3)(−1) |( )| = (3)(−1)|A|

ℎ

| | = (3)(−1). | | = [(3). (−1). (7)] = −21

2 Getting all the parameters for A-1 and A2, then

231

100 −1 1 6 00
6 4 −1),
given = (0 1 1), = (0 21
0 −2 4 0
0
10 0 0)
2 = (0 −1 5 )

0 −10 14

0 0 0

c = (0 ) , = (0

0 −2 4 00

therefore −1 = 1 ( 2 + + )

6

 6=1+c+d

so (-6, 11) satisfy the equation

3 16 2
det = |3 −4 −8|

38 3

16 2
= |0 −2 −4|

0 −10 −3

16 2
7
= −22 |0 1 11 |

0 −10 −3

16 2
7
= −11 |0 1
11|

0 0 27

11

= (−22)(1) (27)

11

= -74

determinant of A ≠0, ⸫matrix A has an inverse.

4 a) Determinant = (3)(4) − (−3)(1) = 12 + 3 = 15

4 3 −43) = (01 10)
(−151 135) (31
15
15

b) Determinant = (5)(−3) − (−2)(4) = −15 + 8 = −7

32
−
(47 − 57) (45 −−23) = (10 01)
7 7

232
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 determinant = 0
Section
k |− 3 22| − 1 |2 0+ 1 22| + 3 |2 0+ 1 − 3|
Clone −6 2 − 2 2 − 4 − 2 + 6 2 + 3 = 0
STPM
4 2 − 7 − 2 = 0
2
(4k +1) (k - 2) =0

= − 1 0r 2

4

| | = 0
1 1
|1 2 3| = 0
1 1
1(2 - 3c) – (1 - 3) + c(c - 2) = 0
2 - 3c + 2 + c2 – 2c = 0
c2 – 5c +4 = 0
(c - 1) (c - 4) = 0
c = 1, c = 4

233 [4 marks]
[4 marks]
3 det M = 1| | − 1 | | + 1 | |
= (ab2 – ac2) – (a2b – bc2) + (a2c – b2c)
= ab2 – b2c + a2c - ac2+bc2 - a2b
=b2(a - c) + ac(a - c)+b(c2-a2)
=-b2(c - a) – ac(c - a) + b(c + a)(c - a)
=(c - a)[-b2 – ac – bc + ab]
= (a - b)[b(c - b)+a(b - c)]
= (a - b)(b - c)(c - a)

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

STPM Past Year

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

234

ANSWER 1(a) 1 −1 1 1 0 0
Section ( | ) = (0 1 −1|0 1 0)
3.1
1 3 20 0 1
STPM
Pass Year 1 −1 1 1 0 0
= (0 1 −1| 0 1 0)

0 4 1 −1 0 1

1 0 1 10
0 1 00|− 1 1 1
= 5 5 5

(0 0 1 − 1 − 4 1
5 5 5)

1 10
111
−1 = − 5 5 5
1 41
(− 5 − 5 5)

1(b) 1 −1 1 1 0 0
( | ) = (0 1 −1|0 1 0)

1 3 20 0 1

1 −1 1 1 0 0
= (0 1 −1| 0 1 0)

0 4 1 −1 0 1

1 −1 1 1 0 0
= (0 1 −1| 0 1 0)

0 0 5 −1 −4 1

1 −1 1 1 0 0
−11|−051 1 01)
= (0 1 4 5
0 0 − 5

1 1 00
0 −1 10|− 1 1 1
= 1 5 5 5

0 0 1 − 1 − 4 1
( 5 5 5)

1 0 1 10
0 1 00|− 1 1 1
= 5 5 5

(0 0 1 − 1 − 4 1
5 5 5)

235
1 10
111

−1 = − 5 5 5
1 41

(− 5 − 5 5)

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

3.1.4 INVERSE MATRICES Quick Notes 3.1.4

1. Notation: Inverse matrix of = −
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.

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

−2 4

236

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

237

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

238

ANSWER 1(a) −3 −1 6 1 0 0
Section ( 2 1 −4 | 0 1 0)
3.1
−5 −1 11 0 0 1
Exercise
3.4 →2 2+ 1→ 1 ( 1 1 −2 1 2 0
2 1 −4 | 0 1 0)

−5 −1 11 0 0 1

→ 2−2 1→ 2 ( 1 1 −2 1 2 0
0 −1 0 | −2 −3 0)

−5 −1 11 0 0 1

→ 3+5 1→ 3 1 1 −2 1 2 0
(0 −1 0 | −2 −3 0)

0 4 1 5 10 1

→− 2→ 2 1 1 −2 1 2 0
(0 1 0 |2 3 0)

0 4 1 5 10 1

→ 3−4 2→ 3 1 1 −2 1 20
(0 1 0 |2 3 0)

0 0 1 −3 −2 1

→ 1− 2→ 1 1 0 −2 −1 −1 0
(0 1 0 |2 3 0)

0 0 1 −3 −2 1

→ 1+2 3→ 1 1 0 0 −7 −5 2
(0 1 0| 2 3 0)

0 0 1 −3 −2 1

−3 −1 6 −1 −7 −5 2

∴ ( 2 1 −4) = ( 2 3 0)

−5 −1 11 −3 −2 1

239

1(b) −4 3 4 1 0 0
( 12 −9 −11 | 0 1 0)

−1 1 1 0 0 1

→ 1⇌ 3 −1 1 10 0 1
( 12 −9 −11 | 0 1 0)

−4 3 4 1 0 0

−→ 1→ 1 1 −1 −1 0 0 −1
( 12 −9 −11 | 0 1 0)

−4 3 4 1 0 0

→ 2−12 1→ 2 ( 1 −1 −1 0 0 −1
0 3 1 | 0 1 12)

−4 3 4 1 0 0

→ 3+4 1→ 3 1 −1 −1 0 0 −1
(0 3 1 |0 1 12 )

0 −1 0 1 0 −4

→ 2⇌ 3 1 −1 −1 0 0 −1
(0 −1 0 |1 0 −4)

0 3 1 0 1 12

→− 2→ 2 1 −1 −1 0 0 −1
(0 1 0 | −1 0 4)

0 3 1 0 1 12

→ 3−3 2→ 3 1 −1 −1 0 0 −1
(0 1 0 | −1 0 4)

00 1 310

→ 1+ 2→ 1 1 0 −1 −1 0 3
(0 1 0 | −1 0 4)

00 1 3 10

→ 1+ 3→ 1 1 0 02 1 3
(0 1 0 | −1 0 4)

001 3 10

−4 3 4 −1 2 13

∴ ( 12 −9 −11) = (−1 0 4)

−1 1 1 3 10

240

1(c) −3 1 21 0 0 → 1⇌ 3 −1 1 10 0 1
(2 3 0|0 1 0) (2 3 0|0 1 0)

−1 1 1 0 0 1 −3 1 2 1 0 0

−→ 1→ 1 ( 1 −1 −1 0 0 −1
2 3 0 |0 1 0)

−3 1 2 1 0 0

→ 2−2 1→ 2 ( 1 −1 −1 0 0 −1
0 5 2 |0 1 2)

−3 1 2 1 0 0

→ 3+3 1→ 3 1 −1 −1 0 0 −1
(0 5 2 |0 1 2)

0 −2 −1 1 0 −3

→ 2+2 3→ 2 1 −1 −1 0 0 −1
(0 1 0 |2 1 −4)

0 −2 −1 1 0 −3

→ 3+2 2→ 3 1 −1 −1 0 0 −1
(0 1 0 |2 1 −4 )

0 0 −1 5 2 −11

→ 1+ 2→ 1 1 0 −1 2 1 −5
(0 1 0 |2 1 −4 )

0 0 −1 5 2 −11

→− 3→ 3 1 0 −1 2 1 −5
(0 1 0 |2 1 −4)

0 0 1 −5 −2 11

→ 1+ 3→ 1 1 0 0 −3 −1 6
(0 1 0| 2 1 −4)

0 0 1 −5 −2 11

−3 1 2 −1 −3 −1 6

∴ ( 2 3 0) = ( 2 1 −4)

−1 1 1 −5 −2 11

241

1(d) 0 1 21 0 0 → 1⇌ 3 1 2 00 0 1
(2 0 1|0 1 0) (2 0 1|0 1 0)

120 001 012 100

→ 2−2 1→ 2 1 2 00 0 1
(0 −4 1|0 1 −2)

0 1 2 10 0

→ 2⇌ 3 1 2 0 00 1
(0 1 2|1 0 0)

0 −4 1 0 1 −2

→ 3+4 2→ 3 1 2 00 0 1
(0 1 2|1 0 0)

0 0 9 4 1 −2

→ 1−2 2→ 2 1 0 −4 −2 0 1
(0 1 2|1 0 0)

0 0 9 4 1 −2

1 1 0 −4 −2 0 1
→9 3→ 3 (0 1 2 | 1 0 0 )
4 1 2
00 1 9 9 − 9

1 0 0 − 2 4 1
0 1 9 9 9
→ 1+4 3→ 1 2 || 1
0 0

(0 0 1 4 1 − 2
9 9 9)

100 − 2 4 1
9 9 9
→ 2−2 3→ 2 | 1 2 4
0 1 0 | 9 − 9 9

(0 0 1 4 1 − 2
9 9 9)

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

( 9 9 − 9)

242

1(e) 2 3 41 0 0 → 1⇌ 3 1 2 40 0 1
(4 3 1|0 1 0) (4 3 1|0 1 0)

124 001 234 100

→ 2−4 1→ 2 1 2 40 0 1
(0 −5 −15 | 0 1 −4)

2 3 4 10 0

→ 3−2 1→ 3 1 2 40 0 1
(0 −5 −15 | 0 1 −4)

0 −1 −4 1 0 −2

−→ 2→ 2 1 2 40 01
(0 −5 −15 | 0 1 −4)

0 1 4 −1 0 2

→ 2⇌ 3 1 2 4 001
(0 1 4 | −1 0 2 )

0 −5 −15 0 1 −4

→ 3+5 2→ 3 1 2 40 0 1
(0 1 4 | −1 0 2)

0 0 5 −5 1 6

→ 1−2 2→ 1 1 0 −4 2 0 −3
(0 1 4 | −1 0 2)

0 0 5 −5 1 6

1 1 0 −4 2 0 −3
→5 3→ 3 (0 1 4 | −1 0 2 )
1 6
00 1 −1 5 5

→ 1+4 3→ 1 1 0 0 −2 4 9
0 1 4 || −1 5 5

0 2

(0 0 1 −1 1 6
5 5)

1 0 0 −2 4 9
5 5
→ 2−4 3→ 2 |
0 1 0 | 3 − 4 − 14
5 5
1 6
(0 0 1 −1 5 5)

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

(−1 5 5 )

243

1(f) 2 −1 −1 1 0 0
(−1 2 0 | 0 1 0)

−1 0 3 0 0 1

→ 1+ 2→ 1 1 1 −1 1 1 0
(−1 2 0 |0 1 0)

−1 0 3 0 0 1

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

−1 0 3 0 0 1

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

01 2 111

→ 2⇌ 3 1 1 −1 1 1 0
(0 1 2 |1 1 1)

0 3 −1 1 2 0

→ 3−3 2→ 3 1 1 −1 1 1 0
(0 1 2|1 1 1)

0 0 −7 −2 −1 −3

→ 1− 2→ 2 1 0 −3 0 0 −1
(0 1 2|1 1 1)

0 0 −7 −2 −1 −3

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

21 3
00 1 77 7

1 0 −3 0 0 −1
35 1
→ 2−2 3→ 3 0 1 0 || 7 7 7

(0 0 1 21 3
7 7 7)

632
100 777
→ 1+3 3→ 1 |
0 1 0 | 3 5 1
7 7 7
213
(0 0 1 7 7 7)

244

632

2 −1 −1 −1 7 7 7
∴ (−1 2 0) = 3 5 1

−1 0 3 777
213

(7 7 7)

2 1 7 3 5 −7 1 800

= (0 5 2 ) (−6 10 2 ) = (0 8 0)

3 0 −1 15 −21 −5 0 0 8

= 8

− = 8 − [Multiply − on the left of both

sides]
= 8 −

5 71
−8
− = 1 = 1 5 −7 1 8 5 8
8 8 (−6 10 2 )= 1
−21 3 4
15 −5 −4 21 4
15 5
( 8 − 8 − 8)

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

= 2

− = 2 − [Multiply − on the left of both

sides]
= 2 −

11
1
− = 1 = 1 1 2 1 2 2
2 2 (0 1 2) = 0 1 1
1 2
0 0 1

(0 2 0)

245

4 = ( + 4 )

124 124 100

= ( 4 3 8 ) [( 4 3 8 ) + 4 (0 1 0)]

−4 −4 −9 −4 −4 −9 001

124 524
= ( 4 3 8 )( 4 7 8 )

−4 −4 −9 −4 −4 −5

−3 0 0
= ( 0 −3 0 )

0 0 −3

= −3

− = −3 − [Multiply − on the left of both

sides]
= −3 −

− = − 1
3

524
−3 −3 −3
− = − 1 = − 1 5 2 4 4 7 8
3 3 (4 7 8 )=
−4 −3 −3 −3
−4 −5 4 4 5

(3 3 3)