Chapter 4 Matrices

LIM HWEE CHENG
MATHEMATICS SM015

CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q1 : 2003/ 2004

(a)Let matrix A  6  4 . If A2  pA  qI 0 where p and q are real numbers, I is a 22 identity
1 
0 

matrix and 0 is a 2 2 null matrix. Find p and q.

(b) Given a matrix equation Ax  B as

1 1  3 x 2
2 4  y 3
1   

1 1 1  z  1

(i) Find the determinant of matrix A.

5 p 3
(ii) Given the cofactor matrix of A   4 2  2 , find p and q.

 q  2 1 

(iii) Determine the adjoint matrix of A and hence find the inverse of A.

(a) Given A2  pA  qI  0
[16 −04] [61 −04] − [16 −04] − [10 10] = [00 00]

[362 −−244] − [6 −04 ] − [0 0 ] = [00 00]

[32 − 6 − −24 + 4 ] = [00 00]
6 − −4 −

Compare −4 − = 0
6 − = 0 = −4
= 6

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

1 1  3 x 2
Given 2 4  y 3
(b) 1   

1 1 1  z  1

(i) ⌈ ⌉ = 1 |−11 −−14| − (−1) |21 −−41| + (−3) |21 −11|
= (1 + 4) + (−2 + 4) − 3(2 + 1)

= −2
(ii) = 12 = (−1)1+2 |21 −−41| = −(−2 + 4) = −2

= 31 = (−1)3+1 |−−11 −−43| = 4 − 3 = 1

 5 2 3 
(iii) Cofactor C   4 2  2

 1  2 1 

5 −4 1
Adj (A) = = [−2 2 −2]

3 −2 1

−1 1 1 5 −4 −5 2 −1
| | −2 2 −1
Thus, = ( ) = [−2 −2 12 1 2
3 −2] = ( 1
1 −3 1)

−1

22

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q2 : 2003/2004

(a) Matrices A and B are given as

 1 2 3  4 1  4
A  1 0 4 , B  1 1 3.5
 0 2 2  1 1 1

Find AB and hence find A1.

(b) A company produces three grades of mangoes: X, Y and Z. The total profit from 1kg
grade X, 2kg grade Y and 3kg grade Z mangoes is RM20. The profit from 4kg grade Z
is equal to the profit from 1kg grade X. The total profit from 2kg grade Y and 2kg
grade Z mangoes is RM10.
(i) Obtain a system of linear equations to represent the given information.
(ii) Write down the system in (i) as a matrix equation.
(iii) Use the inverse method to solve the system of linear equation. Hence, state the profit per kg
for each grade.

(a) 1 2 3 4 1  4 5 0 0 = 1 0 0
AB  1 0 4  1 1 3.5  0 5 0 5 [0 1 0] = 5I

 0 2 2 1 1 1 0 0 5 0 0 1

To find the − , = 5

− = − (5 )

= 5 −

4 −1 −4

4 −1 −4 5 5 5
− 1 1 −1 −1 7
= 5 = 5 [−1 −1 3.5] =
1 1 −1 5 5 10
1 1 −1
[5 5 5]

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

(b) (i) + 2 + 3 = 20
4 = −
2 + 2 = 10

1 2 3 20
(ii) [−1 0 4] [ ] = [ 0 ]

0 2 2 10

(iii) =

− = −

= −

= − 1 4 −1 −4 20
5
= [−1 −1 3.5] [ 0 ]
1 1 −1 10

1 40
5
= [15]
10

8
=[3]

2

Therefore, = 8, = 3 , = 2

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q3 : 2004 / 2005

x2 x 1
 
(a) Show that the determinant of the matrix A   y 2
y 1 is ( y  x)(z  x)( y  z) for
 z 2 z 1


real x, y and z

1 1 1
(b) By substituting x  1, y  2 and z  3 , the matrix A becomes A  4 2 1 .

9 3 1

Find the adjoint and inverse of the matrix A.

(c) The graph of a quadratic equation y  ax 2  bx  c passes through the points whose

coordinates are 1,2 , (2,3) and (3,6) , and

(i) obtain a system of linear equations to represent the given information.

(ii) Write down the system in (i) as a matrix equation in the form of AX  B where

a
X  b

 c 
(iii) Use the inverse of matrix to solve the system of linear equation in (ii). Hence, find the
quadratic equation of the graph

x2 x 1+
 y 1
(a) A   y 2 z 1+

 z 2


Expansion 1st row

| | = 1 | 22 | − 1 | 22 | + 1 | 22 |


= 2 − 2 − [ 2 − 2] + [ 2 − 2]

= ( 2 − 2) − 2( − ) − ( − )

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

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

= ( − )( − )( − ) Shown

1 1 1
(b) Given A  4 2 1

9 3 1

⌈ ⌉ = (2 − 1)(2 − 3)(3 − 1) = (1)(−1)(2) = −2

|32 11| − |94 11| |94 32| = −1 5 −6
Cofactor matric, C = − |31 11| |19 11| − |91 13| [2 −8 6]
− |14 11| |41 21| ] 3
[ |21 11| −1 −2

−1 2 −1
Adj(A) = = = [ 5 −8 3 ]

−6 6 −2

11
−1
−1 = 1 ( ) = 1 −1 2 −1 2 4 2
| | −2 [5 −8 3 ] = −5 −3
−6 6 −2 2
2
[ 3 −3 1 ]

(c) i.
Given y  ax2  bx  c
For (1,2) : + + = 2 − − − − − (1)
(2,3) : 4 + 2 + = 3 − − − − − −(2)
(3,6) ∶ 9 + 3 + = 6 − − − − − −(3)

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

1 1 1 2

ii. (4 2 1) ( ) = (3)

9 3 1 6

AX = B

iii. A-1AX = A-1B

= A-1B

X = A-1B

−1 2 −1 2
1 3 ] [3]
( ) = −2 [5 −8 −2 6
−6 6

1 −2
−2
= [4]
−6

1
= [−2]

3

= 1 , = −2, = 3

Quadratic Equation : = − 2 + 3

Q4 : 2005/2006

− − − = [ − − ]
Given = [ ] ,    = [− ] and
−
−

(a) Find matrix D  A  (BC)T

(b) Show that |AD| = |DA|

(a) = − [ − − ] = −1 −1 1
[− ] [−2 1 0]
−2 1
− 1

− − −1 −1 1 0 2 −3

Thus, = [ ] −   [−2 1 0] = [ 3 0 2 ]

− 1 −2 1 −2 1 −1

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

− − 0 2 −3 4 −4 5
(b) = [ ] [ 3 0 2 ] = [3 4 −4]

− −2 1 −1 3 −2 5
| | = 4 |−42 −54| − (−4) |33 −54| + 5 |33 −42|

= 4(20 − 8) + 4(15 + 12) + 5(−6 − 12)

= 66

0 2 −3 − − 7 −1 0

= [ 3 0 2 ] [ ] = [−5 2 −6]

−2 1 −1 − 504

| | = 7 |20 −46| − (−1) |−55 −46| + 0 |−55 02|
= 7(8 − 0) + (−20 + 30) + 0

= 66

Thus , |AD| = |DA|

Q5 : 2006/2007

− and = [ − ], find matrix R such that
If = [ ] −
−
]


+ ( ) = [−

−

1 −2 [01 0 −01] = 1 2 −1
= [1 1] −1 [1 −1 −1]
−1 0 1 0
0

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

0 22
+ 2( ) = [−2 4 3]

−4 5 3

0 22 1 2 −1

= [−2 4 3] − 2 [1 −1 −1]

−4 5 3 01 0

−2 −2 4
= [−4 6 5]

−4 3 3

Q6 : 2007/2008
1 a 2

Given that A  2 1 2 , where a and b are constants.
2 2 b
4 1 2

a) If A  13 , evaluate the determinant of matrix 2 a 2 using determinant properties.

4 2 b
b) Given that A2  4A  5I , where I is a 3 3 identity matrix. Show that a  2 and b  1. Hence

find A1 .

1a2
(a) Given | | = | 2 1 2 | = −13

22b

4 1 2 2(1) 1 2 1 ↔ 2
| 2 a 2 | = |2(1) 2|

4 2 b 2(2) 2

112
= 2 |1 2|

2 2

1 2
= 2(−1) |1 1 2|

2 2

= −2(−13) = 26

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

(b) A2  4A  5I

1 a 21 a 2 1 a 2 1 0 0
2 1 22 1 2  42 1 2  50 1 0
2 2 b2 2 b 2 2 b 0 0 1

5 + 2 2 + 4 2 + 2 + 2 4 8 5 0 0

[8 2 + 5 6 + 2 ] − [8 4 8 ] = [0 5 0]

6 + 2 2 + 2 + 2 8 + 2 8 8 4 0 0 5

2 + 1 + 4 2 + 2 − 6 5 0 0

[0 2 + 1 −2 + 2 ] = [0 5 0]

−2 + 2 2 + 2 − 6 8 + 2 − 4 0 0 5

Comparing −2 + 2 = 0
2 + 1 = 5 = 1

= 2

A2  4A  5I

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

5

1 122 400
5
= [[2 1 2] − [0 4 0]]
2 2 10 0 4

1[ −3 2 2

= 5 2 −3 2 ]
2 2 −3

−3 2 2

5 55
−3 2
= 2
5 55
2 −3
2
[5 5 5]

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q7: 2008 /2009

1 2 1  2 2  3
(a) Given the matrices P  2 2  
7. 1 and Q   2 1 0  . Find PQ and hence

1 2 2  3 0 3 

determine P 1 .

(b) The following tables shows the quantities (kg) and the amount paid (RM) for the three
types of items bought by three housewives in a supermarket.

Housewives Sugar (kg) Flour (kg) Rice (kg) Amount paid
(RM)

Aminah 3 6 3 16.50

Malini 6 3 6 21.30

Swee Lan 3 6 6 21.00

The price in (RM) per kilogram (kg) of sugar, flour and rice are x, y and z respectively.
i) Form a system of linear equations from the above information and write the

system of linear equations in the form of matrix equation AX  B .
ii) Rewrite AX  B above in the form kPX  B , where A  kP, P is the matrix in

(a) and k is a constant. Determine the value of k and hence find the values of x,
y and z.

1 2 1  2 2 3
PQ  2 2  
(a) 1  2 1 0 

1 2 2 3 0 3 

3 0 0
 0 3 0

0 0 3

100
= 3 [0 1 0]

001

= 3

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

To find the −1

PQ  3I

P1PQ  3P1I
= 3 −1
3 −1 =

P1  1 Q
3

1  2 2 3
3  
Thus, P1   2 1 0 
0
3 3 

2 2 −1

33

= [ 2 −1 0 ]

33

−1 0 1

(c) i. 3 + 6 + 3 = 16.50
6 + 3 + 6 = 21.30
3 + 6 + 6 = 21.00

3 6 3 16.50

(6 3 6) ( ) = (21.30)

3 6 6 21.00

1 2 1 16.50

ii. 3 (2 1 2) ( ) = (21.30)

1 2 2 21.00

=

Thus, = 3

3 =
= 1

3

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

1 2 1 1 16.50
3
(2 1 2) ( ) = (21.30)
1 2 2 21.00

Multiply P-1 to both side to solve for X

− = 1 −
3

= 1 −

3

2 2 −1 16.50
Thus , ( ) = 0 ) (21.30)
1 ( 3 3 1 21.00
3 2 −1

3 3

−1 0

1 4.2
3
= (3.9)
4.5

1.4
= (1.3)

1.5

Therefore, = 1.40 , = 1.30 , = 1.50

Q8 : 2009/2010

3 x 2x 
Matrix A is given as A  0 x 
4  and A  75 . Find

0 0 x 10

a) the value of x.

b) the cofactor and the adjoint matrix of A. Hence, determine the inverse of A.

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

(a)

Given | | = −75

3 2
|0 4 | = −75
0 0 − 10

Expansion of 3rd row

0 | 24 | − 0 |30 24 | + ( − 10) |03 | = −75

0 − 0 + ( − 10)(3 − 0) = −75
3 2 − 30 + 75 = 0
2 − 10 + 25 = 0
( − 5)2 = 0
= 5

3 5 10 
(b) A  0 5 
4 

0 0  5

|05 −45| − |00 −45| |00 50| = −25 0 0
Cofactor = − |50 −105| |30 −105| − |03 50| ( 25 −15 0)
|03 55| ) −12
( |55 140| − |30 140| −30 15

−25 25 −30
Adjoint matrix , ( ) = = ( 0 −15 −12)

0 0 15

1 −1 2
−25 25 −30 3 35

Inverse Matrix −1 = 1 ( ) = 1( 0 −15 −12) = 0 1 4
| | 0 5 25
−75 0 15

(0 0 −1

5)

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q9: 2009/2010

The following table shows the quantities in kilogram (kg) and the amount paid (RM) for three types of
fruits bought from three stalls at a night market.

Stall Fruit Mango Durian Rambutan Amount paid
(kg) (kg) (kg) (RM)
P 5 3 2 34.00
Q 3 4 4 37.00
R 2 3 4 29.00

The price in RM per kilogram (kg) for mango, durian and rambutan are x, y and z respectively.
a) Form a system of linear equations which represent the total expenditure per stall

calculated based on the weight bought and price per kiligram. Hence, write the

system in the form of a matrix equation AX  B .

b) Find the determinant, minor and adjoint of matrix A.
c) Based on part (b) above, find A1 . Hence, solve the matrix equation.

5 + 3 + 2 = 34
(a) 3 + 4 + 4 = 37

2 + 3 + 4 = 24

5 3 2 34

(3 4 4) ( ) = (37)

2 3 4 24

AX = B

(b) | | = 5 |43 44| − 3 |32 44| + 2 |32 43|

= 5(16 − 12) − 3(12 − 8) + 2(9 − 8)
= 10

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

|34 44| |32 44| |23 34| 4 4 1
Matrix Minor , M = |33 42| |52 33| = (6 16 9)
|25 24| 14 11
(|43 24| |35 42| |53 43|) 4

4 −4 1
Matrix Cofactor C = (−6 16 −9 )

4 −14 11

4 −6 4
Adjoint Matrix , Adj (A) = = (−4 16 −14)

1 −9 11

2 −3 2

−1 1 1 4 −6 4 555
| | 10 −2 8 −7
(c) Inverse Matrix = ( ) = (−4 16 −14) =
1 −9 11 555
1 −9 11

(10 10 10)

AX = B

A-1AX = A-1B

IX = A-1B

1 4 −6 4 34
10
X = (−4 16 −14) (37)
1 −9 11 24

3
= (5)

2

Therefore, = 3, = 5, = 2

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q10: 2010/2011

The following table shows the price(RM) per type of 0.5kg cakes old at the shops P, Q and R together
with the total expenditure if a customer buys a number of each type of cake from the listed shop.

Cake types Total
Expenditure
Shops Banana Chocolate Vanilla (RM)
36
P5 8 5
Q4 6 6 30
R5 9 7
40

Let the number of banana, chocolate and vanilla cakes bought from each shop be x, y and z
respectively.

(a) Write the matrix equation AX  B using above information.

(b) Obtain the adjoint matrix of A. Hence, find the inverse of matrix A.
(c) Determine the value of x, y and z using the inverse matrix of A obtained in (b).

5 + 5 + 5 = 36
(a) 4 + 6 + 6 = 30

5 + 9 + 7 = 40

5 8 5x 36
4 6 6 y  30
5 9 7 z  40

(b) Determinant of A

A  6 6 84 6  4 6
5 5
97 57 59

 5(42  54)  8(28  30)  5(36 - 30)

 14

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

6 6 4 6 4 6
  
 9 7 57 5 9   12   2  6   12 2 6

 85 55 5 8    11  10 5    5
Cofactor C  57  9     18      11 10
 9 7 55 8  
 5   10   2  18 10  2
8 5 46 5

 6 6 4 6 
 

12 2 6 T 12 11 18 
Adjoint A  11  5  10 10
10   2

 18 10  2  6  5  2 

A1  1 AdjA
A

12 11 18 
1 10 10
  2  5  2 
 14  6

6 11 9 
 71 7 
14 5 
5 
 7 7 7
 3 5 1
7 14 7 

Q11: 2011/2012

1 2 1

Matrix A is given as 2 3  3.

2 2 1

 3 x  y  2
 
(a) Given the cofactor of A is  0 1 2  where x  0. Determine the values of x and y.

 3 x 2 1

(b) Given A2  4A I  0, show that A3  15 A  4I where I is the 3×3 identity matrix. Hence, find A3 .

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

(a) Given C12  x  y
x  y  (1)12 2  3
2 1
+ = −(−2 + 6)
+ = 4

y  x  4      (1)

Given 32 = 2

 1 32 1 1  x2

23

x2 1

x  1 (Given x > 0 )

From Eq (1) : y  1 4  5

(b) Given A2  4 A  I  0
A2  4A  I Multiply A both sides

A A2  A 4A  I 

3 = 4 2 −
= 4(4 − ) −
= 16 − 4 −

A3  15 A  4I shown

1 2 1 1 0 0 11 30 15
(c) A3  152 3  3  40 1 0  30 41  45

2 2 1 0 0 1 30 30 19

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q12:2011/2012

The following table shows the quantities (unit) and the amount paid (RM) for pens bought from three

shops.

Pilot Kilometrico Papermate Amount paid
(unit) (unit) (unit) (RM)

S 1 p 2p 18.00
T 1 q 3q 31.00
U 1 r 4r 37.00

Given the price in RM per unit of pilot, kilometrico and papermate pens by x, y and z respectively.

(a) Obtain a system of linear equations to represent the given information.

(b) Write the system in the form of matrix equation AX  B where X  x 
  y .
 z 

(c) Given the minor a11 , a21 and a22 of matrix A is 9, 12 and 8 respectively. Find the values of p,

q and r.

(d) Find the determinant, cofactor, adjoint and A1 of matrix A. Hence, find the values of x, y

and z.

+ + 2 = 18
(a) + + 3 = 31

+ + 4 = 37

1 2 18

(b) (1 3 ) ( ) = (31)

1 4 37

AX = B

(c) Given 11 = 9
| 34 | = 9

4 − 3 = 9

= 9 − − − − − −(1)

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Given 21 = 12
| 24 | = 12

4 − 2 = 12

= 6 − − − − − −(2)

Given 22 = 8
|11 24 | = 8

4 − 2 = 8

= 2 − 4 − − − − − (3)

Substitute Eq(3) into Eq(2)
(2 − 4) = 6
2 2 − 4 − 6 = 0
2 − 2 − 3 = 0
( − 3)( + 1) = 0
≠ −1 ( > 0 because is quantity unit ) @ = 3

From Eq(3) , = 2(3) − 4
From Eq(1) , = 2
(3) = 9  = 3

1 2 4 18

(d) (1 3 9 ) ( ) = (31)

1 3 12 37

A X =B

Determinant of Matrix A ( Expansion of 1st row ) 21
| | = +1|33 192| − 2 |11 192| + 4 |11 33|

= (36 − 27) − 2(12 − 9) + 4(3 − 3)
=3

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Matrix Cofactor C

|33 192| − |11 192| |11 33| = 9 −3 0
= − |32 142| |11 142| − |11 23| (−12 8 −1)
− |11 49| |11 23| ) −5 1
( |23 94| 6

9 −12 6
( ) = = (−3 8 −5)

0 −1 1

9 −12 6 3 −4 2
8 −5)
Inverse Matrix −1 = 1 ( ) = 1 (−3 −1 = (−1 8 − 5
3 0
| | 3 3)
−1
10 1

33

From A X = B

A-1 A X = A-1 B

IX = A-1 B

1 9 −12 6 18
3
X = = A-1 B = (−3 8 −5) (31)
0 −1 1 37

1 12
3
= ( 9 )
6

4
= (3)

2

Thus, = 4 , = 3, = 2

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q13 : 2012 / 2013

 3 1
(a) Matrix M is given as  4 . Show that M2 where I is the 2×2 identity matrix. Deduce
4   7M 8I,

M 1  7 I  1 M .

that 8 8

p 1 1 1 
 
(b) Given matrix A   3 2 4  and A  27. Find the value of p, where p is an integer.

 1 0 p  2

(a) M  3  1
 4 
4 

2 = [−34 −41] [−34 −41] = [−1238 −207]

7 − 8 = 7 [−34 −41] − 8 [01 10] = [−1238 −207]
M 2  7M 8I, shown

To deduce the M-1

M 2  7M  8I
MMM 1  7MM 1  8IM 1
M  7I  8M 1
8M 1  7I  M
M 1  7 I  1 M

88

p 1 1 1 
 
(b) A  3 2 4 
 2
+ - +

1 0 p

Given | | = 27 ( Expansion of 3rd row )

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

1 1 p 1 1  p  2 p 1 1
1 0  27
24 3 4 32

1 6 p  22p  2  3  27
6  p  22p  5  27

6  2 p2  9 p 10  27

2 p2  9 p 11  0

2p 11p 1  0

p  11 ignore  or p  1 ( Given p is integer )

2

Q14: 2012 / 2013

2 2 3

Given A  1 5 4

3 1 4

(a) Find the determinant of matrix A.

(b) Find the minor, cofactor and adjoint of matrix A.

(c) Given A(Adjoint (A))  A I, where I is 3×3 identity matrix, show that A1  1 adjo int A.
A

Hence, find A1 .

(d) By using A1 in part (c), solve the following simultaneous equations.

2x  2 y  3z  49
x  5y  4z  74
3x  y  4z  49

(a) Determinant of matric A ( Expansion of 3rd row )

| | = 3 |25 34| − 1 |12 34| + 4 |12 52|
= 3(8 − 15) − 1(8 − 3) + 4(10 − 2)
=6

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

|15 44| |31 44| |13 15| 16 −8 −14
(b) Minor M = |21 34| |32 43| |32 12| =( 5 −1 −4 )
34| |21 43| |21 52|) 5 8
(|52 −7

+(16) −(−8) +(−14) 16 8 −14

Cofactor C = ( −(5) +(−1) −(−4) ) = (−5 −1 4 )

+(−7) −(5) +(8) −7 −5 8

16  5  7
Adjoint Matrix Adj(A) = = ( 8 1  5 )

14 4 8

(c)

Aadjoint A  A I

A1 Aadjoint A  A1 A I
I adjoint A  A A1I
adjoint A  A A1
A1  1 adjoint A

A shown

A1  1 AdjA
A

1  16  5  7
6   5
  8 1 8 
4
 14

8 5  7 
 6  
 3 6 
 4 1 5 

 3 6 6
 7 2 4
3 3 3 

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

(d)

2x  2 y  3z  49
x  5y  4z  74
3x  y  4z  49

2 2 3 49

(1 5 4) ( ) = (74)

3 1 4 49

AX=B

Multiply A-1 both sides

A-1 A X = A-1 B

IX = A-1 B

X = A-1 B

1  16  5  749
6   574
  8 1 8 49
4
 14

1 71
6
= [73]
2

71

6
= 73

6
1
[3]

x  71, y  73 , z  1
6 63

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q15: 2013/2014

5 3 0 and c 3 0
1 0 B  3 d 0. Find the values of c, d and e such that AB  14I,
Given matrices A  3
0 0 2 0 0 e

where I is the identity matrix. Hence, determine A1.

Given AB 14I

5 3 0c 3 0 1 0 0
3 1 03 d 0  140 1 0
0 0 20 0 e 0 0 1

5 + 9 15 + 3 0 14 0 0
[3 − 3 9 − 0 ] = [ 0 14 0 ]

0 0 2 0 0 14

Comparing : 5 + 9 = 14 9 − = 14 2 = 14
= 7
5 = 5 = −5

= 1

From AB 14I

A-1AB = 14 A-1 I

IB = 14 A-1

1 30
130 14 14

A-1 = 1 = 1 [3 −5 0] = 3 −5 0
14 14 0 0 7 14 14
1
[0 0
2]

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q16: 2013/2014
An osteoporosis patient was advised by a doctor to take enough magnesium, vitamin D and calcium to
improve bone density. In a week, the patient has to take 8 units magnesium, 11 units vitamin D and 17
units calcium. The following are three types of capsule that contains the three essential nutrients for the
bone:

Capsule of type P: 2 units magnesium, 1 unit vitamin D and 1 unit calcium.
Capsule of type Q: 1 unit magnesium, 2 units vitamin D and 3 units calcium.
Capsule of type R: 4 units magnesium, 6 units vitamin D and 10 units calcium.

Let x, y and z represent the number of capsule of types P, Q and R respectively that the patient

has to take in a week.

(a) Obtain a system of linear equation to represent the given information and write the system

x
 y.
in the form of matrix equation AX  B, where X  

 z 

(b) Find the inverse of matrix A from part (a) by using the adjoint method. Hence, find the values

of x, y and z.

(c) The cost for each capsule of type P, Q and R are RM10, RM15 and RM17 respectively. How
much will the expenses be for 4 weeks if the patient follows the doctor’s advice?

(a) System of linear equation System of matrix equation

2 + + 4 = 8 2 1 4 8
+ 2 + 6 = 11
+ 3 + 10 = 17 (1 2 6 ) ( ) = (11)

1 3 10 17

(b) | | = 2 |32 160| − 1 |11 160| + 4 |11 23| = 2(20 − 18) − (10 − 6) + 4(3 − 2) = 4

+ |23 160| − |11 160| + |11 23| = 2 −4 1
Cofactor = − |31 140| + |21 140| − |21 31| (2 16 5)
− |12 46| + |12 21|) −8 3
( + |21 64| −2

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

2 2 −2
( ) = = (−4 16 −8)

1 −5 3

2 1 1 −1
2 −2 2
Inverse Matrix −1 = 1 ( ) = 1 (−4 16 −8) = (−1 2 2
| | 4 1 −5 31
4 −2)
−4
3

4 54

2 1 4 8

From (1 2 6 ) ( ) = (11)

1 3 10 17

AX = B

Multiply A-1 both sides

A-1 A X = A-1 B

IX = A-1 B

X = A-1 B

1 1 −1

2 2 2 8 1
( ) = (−1 4 −2) (11) =(2)

1 − 4 3 17 1

4 54

Thus, = 1 , = 2, = 1

c) The expanses per 4 week = 4 × [10(1) + 15(2) + 17(1)] = 228

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q17:2014/2015

1 0 0 1 0 0
Given matrices A  4 0  0 where B is the inverse of A. find x, y and z in terms of
1 and B  z 1

a b 1 x y 1

a and b.

Given that B is the inverse of A, thus =

1 0 0
[ ] [ ] = [0 1 0]
0 0 1

1 0 0 100

[ 4 + 1 0]=[0 1 0]

+ + + 1 0 0 1

Comparing : + = 0 + + = 0
4 + = 0 = − + (−4) + = 0
= −4
= 4 −

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q18:2014/2015

Two companies P and Q decided to award prizes to their employees for three work ethical values,
namely punctuality (x), creativity (y) and efficiency (z). Company P decided to award a total RM3850
for the three values to 6, 2 and 3 employees respectively, while company Q decided to award RM3200
for three values to 4, 1 and 5 employees respectively. The total amount for all the three prizes is RM1000.

(a) Construct a system of linear equations to represent the above situations.
(b) By forming a matrix equation, solve this equation system using the elimination method.
(c) With the same total amount of the money spent by company P and Q, is it possible for

company P to award 15 employees for their creativity instead of 2 employees? Give your
reason.

(a) System of Linear Equations

6 + 2 + 3 = 3850
4 + + 5 = 3200

+ + = 1000

(b) System of Matrix Equations

6 2 3 3850

(4 1 5) ( ) = (3200)

1 1 1 1000

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Elimination Method ( Gauss Jordon Elimination Method )

6 2 3 3850 ∗ = −
(4 1 5|3200) ∗ ∗= = −−

1 1 1 1000

6 2 3 3850 ∗ = +
( − | ) ∗ = +



∗ = −
( − | ) ∗ = −



1∗ = 1 1 2∗ = −1 2 3∗ = 1 3
( − |− ) 234 78 39



Thus, = , = , =
( | )



(c)

Company P 6 3 3850

Company Q (4 1 5) ( ) = (3200)
1 1 1
1000

∗ =

| ∗| = 6 |11 15| − 15 |14 15| + 3 |14 11|
= 6(1 − 5) − 15(4 − 5) + 3(4 − 1)

=0

Since | ∗| = 0 , ∗ is a singular matrix. Therefore, inverse matrix does not exist. The system will
have no solution.

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q19: 2014/2015

−
(a) Determine the values of x so that [ ] is singular.

−

3 5 1 4 
(b) If A  1 2 and B  0 1, find C when A  BCB 1.

(a) Singular Matrix, | | = 0
| T yp e+ equ−a  ti o|n=he0re.
−

Expansion of 2nd row
− | 3 −−11| + 0 |11 −−11| − 1 |11 3 | = 0
− (− + 3) + 0 − 1(3 − ) = 0

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

(b) Given = −1

= −1

=

− = −

− =

= −

= 1 [−01 −14] [13 25] [01 −41]
−1

= [−71 −132] [01 −41]

= [−71 −152]

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q20: 2015/2016

A curve y  ax2  bx  c where a, b and c are constants, passes through the points (2,11), (1,16) and

(3,28).
(a) By using the above information, construct a system containing three linear equations.
(b) Express the above system as a matrix equation AX  B.
(c) Find the inverse of matrix A by using the adjoint matrix method. Hence, obtain the values of
a, b and c.

(a)

Given y  ax2  bx  c

At point ( 2, 11 ) : (2)2 + (2) + = 11 ⇒ 4 + 2 + = 11
At point (1,16) : (−1)2 + (−1) + = −16 ⇒ − + = −16
At point (3,28) : (3)2 + (3) + = 28 ⇒ 9 + 3 + = 28

(b) System of Matrix Equation

4 2 1 11

(1 −1 1) ( ) = (−16)

9 3 1 28

AX = B

(c) | | = +4 |−31 11| − 2 |19 11| + 1 |19 −31|
= 4(−1 − 3) − 2(1 − 9) + 1(3 + 9)
= 12

|−31 11| − |19 11| |91 −31| −4 8 12
Cofactor = − |23 11| |94 11| − |94 32| =( 1 −5 6)
11| − |41 11| |41 −21|) −3 −6
( |−21 3

−4 1 3
( ) = = ( 8 −5 −3)

12 6 −6

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

−1 1 1

−4 1 3 3 12 4

Inverse Matrix −1 = 1 ( ) = 1 ( 8 −5 −3) = 2 −5 −1
| | 12 12 6 −6 3
12 4
( 1 1 −1

2 2)

AX = B

Multiply A-1 both sides

A-1 A X = A-1 B

IX = A-1 B

X = A-1 B

−4 1 3 11 24 2
( )= 1 1
12 ( 8 −5 −3) (−16) = 12 ( 84 ) = (7)
12 6 −6 28 −132 −11

Thus, = 2, = 7 & = −11

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q21 : 2016/2017

(a) Given = [− − ], = − Find R 1 by using elementary row operation
[
− ] and = [ ].



method. Hence , if RX  3Q  PT , determine the matrix X.

(b) Ahmad bought an examination pad, 2 pens and a tube of liquid paper for RM 18. Ali spent RM 24
for 3 examination pads, 2 pens and 2 tubes of liquid paper. In the meantime Abu spent RM 36 at the
same store for 3 examination pads, 4 pens and a tube of liquid paper. Let x , y and z represent the price
per unit for examination pad, pen and liquid paper respectively.

(i) Obtain the system of linear equations from the above information.

(ii) Write the system in the form of matrix equation AX  B .

(iii) State the price of each unit of examination pad, pen and liquid paper.
(iv) Aminah bought 4 examination pads, 5 pens and 1 tube of liquid paper. What is the

total price paid?
(a)

Elementary Row Operation

1 2 11 0 0 2∗ = 2 − 3 1
(3 2 2|0 1 0) 3∗ = 3 − 3 1

3 4 10 0 1

1 2 11 0 0 1∗ = 2 1 + 2
(0 −4 −1|−3 1 0)
0 −2 −2 −3 0 1 3∗ = 2 3 − 2

2 0 1 −1 1 0 1∗ = 3 1 + 3
(0 −4 −1|−3 1 0) 2∗ = 3 2 − 3

0 0 −3 −3 −1 2

6 0 0 −6 2 2 1∗ 1 2∗ 1 3∗ 1
6 −12 −3
(0 −12 0 |−6 4 −2) = 1 , = 2 , = 3
0 0 −3 −3 −1 2

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

−1 1 1
100 33

0 1 0|| 1 −1 1
2 36
0 0 11
( 1 −2

3 3)

−1 1 1

33

Thus, −1 = 1 −1 1
2 36

(1 1 −2

3 3)

= +

− = − [ + ]

= − [ + ]

−1 1 1
33 − 2 −1

= 1 −1 1 [3 ( ) + ( 0 6 )]
2 3 6
− −4 2
(1 1 −2

3 3)

−1 1 1
3 3 −1 8
1 −1 1
= 2 (0 12)
3 6 −22 17
(1 1 −2

3 3)

− 19 5

3 3
17
= − 25
6
6 2
41
3)
(3

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

(b) (i) System of linear equations

+ 2 + = 18
3 + 2 + 2 = 24
3 + 4 + = 36

(ii) System of matrix Equation

18

( ) ( ) = (24)

36

=

A=R ∴ − = −

(iii) Multiply A-1 both sides

A-1 A X = A-1 B

IX = A-1 B

X = A-1 B

−1 1 1 18 2

33

( ) = 1 −1 1 (24) = (7)
2 3 6
1 −2 36 2
(1
3 3)

Thus, = 2, = 7 & = 2

2
(iv) Total price = (4 5 1) (7) = [4(2) + 5(7) + 1(2)] = [45]

2

Total price = RM 45

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q22 : 2017/2018

Given matrix = [− ]such that A2  A  I  0 ,  and  are constants, where I and 0 are identity
matrix and zero matrix of 2 2 respectively. Determine the values of  and  .

A2  A  I  0

[−22 35] [−22 35] + [−22 35] + [01 10] = [00 00]

[−−124 1291] + [−22 53 ] + [ 0 0 ] = [00 00]

[−−2 1+42− 2+ 21 + 3 ] = [00 00]
19 + 5 +

Comparing :

− − =
= −
= −

− + (− ) + =
=

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q23: 2017/2018

Given the system of linear equations as follow:
2x  4 y  z  77
4x  3y  7z  114
2x  y  3z  48

 x 

(a) Express the system of equations in the form of matrix equation AX  B where X   y  .

 z 
Hence, determine matrix A and matrix B .

(b) Based on part 23 (a) , obtain A . Hence, find
(i) P if PA  I where I is an identity matrix 3 3.
(ii) Q if Q  (2 A)T

(iii) find adjoint A . Hence, obtain A1 and find the values of x, y and z.

2 4 1 77

(a) (4 3 7) ( ) = (114)

2 1 3 48

AX = B

241 77

Thus, = (4 3 7) and = (114)

213 48

(b) | | = 2 |31 37| − 4 |24 73| + 1 |24 31|
= 2(9 − 7) − 4(12 − 14) + 1(4 − 6)

= 10

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CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

i. Given PA  I , is inverse of A.

Thus, = −

Thus, | | = =
| |

ii. Given Q  (2 A)T

| | = |(2 ) |
= 2| |

= 2(10)

= 20

iii. |13 37| − |24 73| |24 13| 2 2 −2
= − |41 13| |22 31| − |22 41| = [−11 4 6]
− |24 71| |24 43| ] −10 −10
[ |34 17| 25

2 −11 25
( ) = = [ 2 4 −10]

−2 6 −10

Inverse Matrix −1 = 1 ( )
| |

1 −11 5

2 −11 25 5 10 2

= 1 (2 4 −10) = 1 2 −1
10 −2 6 5
−10 5

−1 3 −1)
5
(5

LIM HWEE CHENG 41

LIM HWEE CHENG
MATHEMATICS SM015

CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

AX = B

A-1AX = A-1B

IX = A-1B

X = A-1B

2 −11 25 77
1
( ) = 10 (2 4 −10) (114)
−2 6 −10 48

1 100
10
= (130)
50

10
= (13)

5

= 10, = 13, = 5

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LIM HWEE CHENG
MATHEMATICS SM015

CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q24: 2018/2019

 1 3 4

(a) Given the matrix a  2b 3 2 such that M11  7 and C12  1 , calculate the values

 4 a  b 9

of a and b.

−

(b) Let = [ ], = [− − ] and = [ ]

−

(i) Find determinant of A by expanding first column.

 (ii) Evaluate A2  BT C

(a) = | 3 92| = 7
+

27 − 2( + ) = 7

2 + 2 = 20

+ = 10 − − − − − −(1)

12 = (−1)1+2 | + 2 29| = −1
4

a. (9( + 2 ) − 8) = −1
−(9 + 18 − 8) = −1
9 + 18 = 9
+ 2 = 1 − − − − − − − (2)

Eq(1) – Eq(2) : = −9
= 10 − (−9) = 19

LIM HWEE CHENG 43

LIM HWEE CHENG
MATHEMATICS SM015

CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

1 3 4
(b) i. Given A  1 3 2

4 10 9

Expansion of 1st column
| | = 1 |130 29| − 1 |130 94| + 4 |33 42|

= 1(27 − 20) − (27 − 40) + 4(6 − 12)
= 7 + 13 − 24
= −4

− − 1
ii. ( 2 − ) = [( ) ( ) − ( − )] (1)

− 2

20 52 46 − − 1

= [(12 32 28 ) − ( − )] (1)

50 132 117 − 2

13 53 48 1
= (−1 39 26 ) (1)

56 130 117 2

162
= ( 90 )

420

LIM HWEE CHENG 44

LIM HWEE CHENG
MATHEMATICS SM015

CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

Q25: 2019/2020

 2 3 0
Given matrix A  5 0 4

 0 2 1
(b) Find the determinant of A by expanding first row.

(c) Calculate the adjoint of matrix A. Hence, find A1 .

1

(d) Solve the equation AX  B , where B  2 by using the answer obtained in part (b).

2

(a) Expansion 1st Row
| | = 2 |02 14| − 3 |−05 41| + 0 |−05 02|
= 2(0 − 8) − 3(−5 − 0) + 0(−10 − 0)
= −1

|20 41| − |−05 41| |−05 02|
(b) Matrix Cofactor = − |23 10| |02 01| − |02 23|

( |30 40| − |−25 04| |−25 03|)

+(0 − 8) −(−5 − 0) (−10 − 0)
= (−(3 − 0) (2 − 0) −(4 − 0) )

(12 − 0) −(8 − 0) (0 − (−15))

−8 5 −10
= (−3 2 −4 )

12 −8 15

LIM HWEE CHENG 45

LIM HWEE CHENG
MATHEMATICS SM015

CHAPTER 4 : MATRICE & SYSTEM OF LINEAR EQUATIONS

−8 −3 12
Adjoint Matrix, ( ) = = ( 5 2 −8)

−10 −4 15

Inverse Matrix −1 = 1 ( )
| |

1 −8 −3 12
−1
= (5 2 −8)
−10 −4 15

8 3 −12
= (−5 −2 8 )

10 4 −15

(c) AX  B

A-1AX = A-1B

IX = A-1B

X = A-1B

8 3 −12 1 8 + 6 − 24 −10

( ) = (−5 −2 8 ) (2) = (−5 − 4 + 16) = ( 7 )

10 4 −15 2 10 + 8 − 30 −12

Thus, = −10 , = 7 and = −12

LIM HWEE CHENG 46