Semester 1 Mathematics STPM (Student's Edition)

196

7 6 5
− 16 − 16
16
Hence, P-1 = 5 2 1
− 16
13 16 16
2 7
( 16 16 )
− 16

1 −7 6 −5
= (−5 2 1)
16 13 −2 7

Rewriting the simultaneous equations in matrix form,
1 −2 1
1

( 3 1 2) ( ) = (4)

−1 4 1 2

1
A ( ) = (4)

2

1
( ) = −1 (4)

2

1 −7 6 −5 1
= (−5 2 1 ) (4)
16 13 −2 72

7

16
=5

16
19
(16)

Hence, = 75 19
16 , = 16 and = 16

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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EXAMPLE 3: Solving systems of Linear Equations by Gaussian Elimination Method
A system of linear equations is given by

7x + 16 y − 10z = 9 , 5x + 11y − 8z = −6 , x + y = z

Reduce the augmented matrix of the system of linear equations to row-echelon form and then solve it.

Solution: → x+y−z=0
7x + 16 y − 10z = 9 7x +16 y −10z = 9
5x +11y − 8z = −6
5x + 11y − 8z = −6

x+ y=z

1 1 − 1 0  ⎯⎯⎯⎯⎯→7R1 − R2 → R2 1 1 − 1 0 
7 16 − 10 9  5R1 − R3 → R3 0 − 9 3 − 9
   
 5 11 −8 − 6   0 −6 3 6 

⎯⎯⎯⎯⎯→2R2 −3R3 →R3 1 1 − 1 0 
0 −9 3 −9 
 
 0 0 − 3 − 36  REF

→Solve row 3 0x + 0y − 3z = −36  z = 12

→Solve row 2 0x − 9y + 3z = −9  y = 5

→Solve row 1 x+ y−z=0 x=7

EXAMPLE 4
A system of linear equations is given by,

2x + 3y + 4z =1 , 3x − 2y − 2z = 3 , 4x + 4y + 3z = 5

Reduce the augmented matrix of the system of linear equations to row-echelon form and then solve it.

Solution:

2x + 3y + 4z =1   2 3 4 1  2 3 4 1 
 3 −2 − 2 3
⎯⎯⎯⎯⎯→3x − 2y − 2z = 3 
3R1 − 2R2 → R2  0 13 16 − 3
 3 5 2R1 − R3 → R3  5 − 3
  
4x + 4y + 3z = 5 4 4 0 2

⎯2⎯R2 ⎯− 13⎯R3 →⎯R3⎯→  2 3 4 1 
0 13 16 − 3 

 0 0 − 33 − 33 REF


→Solve row 3 0x + 0y − 33z = 33  z = −1

→Solve row 2 0x +13y +16z = −3  y = 1

→Solve row 1 2x + 3y + 4z = 1  x =1

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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EXAMPLE 5
Solve the following system of equations using augmented matrices:

x + y – z = −2
2x – y + z = 5
−x + 2y + 2z = 1

Solution:
The augmented matrix for this system is

1 1 −1 −2
( | ) = ( 2 −1 1 | 5 )

−1 2 2 1

⎯⎯⎯⎯⎯→R2 − 2R1 → R2 1 1 −1 −2
R1 + R3 → R3 (0 −3 3 | 9 )

0 3 1 −1

⎯R⎯2 +⎯R3 →⎯R3⎯→ 1 1 −1 −2
(0 −3 3|9)

0 0 48

The matrix is now of the row-echelon form. Converting back to a system of equations, we have

x + y – z = −2
−3y + 3z = 9

4z = 8

Hence, z = 2.

Substituting into the first two equations,
y=z–3
=2–3
= −1

and
x = −y + z −2
= (−1) + 2 – 2
=1

Hence, the solution to the system of linear equations is the ordered triple (1, −1, 2).
The system of equations is said to be consistent and has the unique solution.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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EXAMPLE 6

Solve the following system of equations by applying the Gaussian elimination :
2x + y – z = 8

−3x – y + 2z = −11

−2x + y + 2z = −3

Solution:

21 −1 8
The augmented matrix is = (−3 −1 2 |−11)

−2 1 2 −3
.

.

.
1 0 02
= (0 1 0| 3 )

0 0 1 −1

Putting back into equation form, we have x = 2, y = 3, z = −1.
Hence, the solution set for the system of linear equations is ( 2 , 3, −1 ).
The system of equations is said to be consistent and has the unique solution.

EXAMPLE 7
Solve the following system of equations :

x + y – z = −1
2x + 3y + z = 6
5x + 7y + z = 13

Solution:
Using Gaussian elimination,

1 1 −1 1
(2 3 1 | 6 )

5 7 1 13

⎯⎯⎯⎯⎯→−2R1 + R2 → R2 1 1 −1 1
(0 1 3 |4)

0 2 68

⎯−⎯2R2⎯+ R⎯3 →⎯R3⎯→ 1 1 −1 1
(0 1 3 |4)

0 0 00

x + y – z = 1 ( 3 variables )
y + 3z = 4 ( 2 variables )

There is an equation with 3 variables. Hence, there is an infinite number of solutions. The system of
equations is said to be consistent and has infinite or many solutions.

By letting z = t, y = 4 – 3t and x = 4t – 3.
Hence, the solution are x = 4t – 3, y = 4 – 3t, z = t.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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EXAMPLE 8
Solve the following system of equations using Gaussian elimination :

3x −4y + 4z = 7

x − y −2z = 2

2x − 3y + 6z = 5

Solution :

The augmented matrix for the system is
3 −4 4 7

(1 −1 −2|2)
2 −3 6 5

⎯⎯R1 ⎯R2⎯→ 1 −1 −2 2
(3 −4 4 |7)

2 −3 6 5

⎯⎯⎯⎯⎯→R2 − 3R1 → R2 1 −1 −2 2
R3 − 2R1 → R3 (0 −1 10|1)

0 −1 10 1

⎯⎯R2 −⎯R3 ⎯→ R⎯3 → 1 −1 −2 2
(0 −1 10 |1)

0 0 00

Row 3 represents the equation 0 = 0, which is always true. This indicates that the system is dependent
and has infinite number of solutions.

Writing back into equation form, we have
x − y − 2z = 2 …………………… (1)
−y + 10z = 1
y = 10z – 1.…………………(2)

substituting (2) into (1) : x – (10z – 1) – 2z = 2
x – 10z + 1 – 2z = 2
x = 12z + 1

Hence, can be any real number , where ∈ .the solution of the system is the ordered triple of the
form (12t + 1, 10t – 1, t), where ∈ .

Therefore, the system of linear equations has infinitely many solutions.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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EXAMPLE 9
Solve the following system of equations using Gaussian elimination :

2x − 2y − z = 2

4x − 4y − 3z = 2

x−y=5

Solution :
The augmented matrix for this system is

2 −2 −1 2
(4 −4 −3|2)

1 −1 0 5

⎯⎯⎯⎯⎯→R2 − 2R1 → R2 2 −2 −1 2
R1 − 2R3 → R3 (0 0 −1|−2)

0 0 −1 −8

⎯⎯⎯⎯→R3 −R2 →R3 2 −2 −1 2
(0 0 −1|−2)

0 0 0 −6

The third row says that 0x + 0y + 0z = −6, i.e. 0 = −6 which is not true.
This indicates that the system of equations has no solutions.

EXAMPLE 10
A system of linear equations is given by

2x + y − z = 2

3x + 2y − 3z = 2

7x + 4 y + kz = 6

where k is a real number. Write its augmented matrix.
Find the value of k if the system of linear equations has no unique solutions.
With this value of k , reduce the augmented matrix for the system to row-echelon form.
Suggest a general solution for the system of linear equation.

Solution:

 2 1 − 1 2
3 2 − 3 2
 k 6
 7 4

 2 1 −1
No unique solution: Det  3 2 − 3 = 0

7 4 k 

2 −3 −13 − 3 + (−1)3 2 =0
2 74
4k 7k

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2(2k +12)−1(3k + 21)−1(12 −14) = 0

k = −5

⎯⎯⎯⎯⎯→2 1 −1 2 1 1 − 2 0
 3 2 −3 2  R2 − R1 → R1 3 2 −3 2
 
 7 4 −5 6   − 5 6
 7 4

⎯⎯⎯⎯⎯→3R1 − R2 → R2 1 1 − 2 0 
7R1 − R3 → R3 0 1 −3 − 2 
0 3 − 9 − 6

⎯⎯⎯⎯⎯→3R2 −R3 →R3 1 1 − 2 0
0 1 −3 − 2
 0 
 0 0 0

Let z = t , y − 3z = −2  y = 3t − 2

x + y − 2z = 0  x = 2−t

EXAMPLE 11
A systems of linear equations is given by

2x − 2y + z = 1 , 6x + ky − 2z = 5 , 4x + 3y − 3z = 4k , where k is a constant.

a) Write the augmented matrix for the system above and reduce it to row-echelon form.

b) Determine the values of k such that the system has a unique solution.
Hence, find the solutions of the system for the case k = −6 .

c) Determine the values of k such that the system has infinitely many solutions. Hence, find the
solutions of the system.

Solution:

2x − 2y + z =1   2 −2 1 1 
a) 6x + ky − 2z = 5  6 k −2 5 

4x + 3y − 3z = 4k   
 4 3 − 3 4k 

⎯⎯⎯⎯⎯→R2 − 3R1 → R3  2 − 2 1 1 
R3 − 2R1 → R2 0 7 − 5 4k − 2
 
 0 k +6 −5 2 

⎯(⎯k +⎯6)R⎯2 − ⎯7R3 ⎯→ R⎯3 →  2 −2 1 1 
 0 7 −5 4k − 2 

 0 0 5 − 5k 4k 2 + 22k − 26  REF
 

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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b) For unique solution: 5 − 5k  0  k  1

For k = −6 ,

→Row 3 0x + 0y + 35z = −14  z = − 2
5

→Row 2 0x + 7 y − 5z = −26  y = −4

→Row 1 2x − 2y + z = 1  x = − 33
10

c) For infinitely many solutions:

5 − 5k = 0 and 4k 2 + 22k − 26 = 0

k =1 2(k −1)(2k +13) = 0  k =1

k = 1 and k = − 13
2

Let z = t

→Row 2 0x + 7 y − 5z = 2  y = 5t + 2
7

→Row 1 2x − 2y + z = 1  x = − 3t + 11
14

Solutions: x = − 3t + 11 , y = 5t + 2 , z = t, where t  .
14 7

EXAMPLE 12
Solve the system of equations :

2x + y + z = 2
8x + 3y + 5z = 4
3x + y + 2z = -2

Solution:
Using Gaussian elimination,

2 1 12
(8 3 5| 4 )

3 1 2 −2

⎯⎯⎯⎯⎯⎯→−4R1 + R2 → R2 2 1 12
2R3 → R 3 (0 −1 1|−4)

6 2 4 −4

⎯−⎯3R1⎯+ R⎯3 →⎯R 3⎯→ 2 1 12
(0 −1 1| −4 )

0 −1 1 −10

⎯−⎯R 2 ⎯+ R⎯3 →⎯R3 → 2 1 12
(0 −1 1|−4)

0 0 0 −6

From the third row of the matrix, 0x + 0y + 0z = -6
The system of equations has no solution and is said to be inconsistent.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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Hence, given any system of linear equations, there are three possibilities.
1. An independent system has exactly onesolution. There is a unique solution.
2. A depedent system has infinitely many solutions.
3. An inconsistent system has no solution.

EXERCISE 3.5

1. Reduced the following augmented matrices to row-echelon form, and determine

whether this system of linear equations has a unique solution, infinitely many

solutions or no solutions.

Find the solution if exists.

1 − 2 −1 − 8   4 − 2 4 2 1 2 3 3
(a)  3 −1 − 4 −15 (b)  2 −1 − 4 4 (c)  2 3 8 4
   0 3  
 0 1 2 4   6 −3  3 2 17 1 

1 1 − 1 0  4 − 4 3 5   2 1 1 2 
(d)  2 3 1 0 (e) 1 − 2 1 3  (f) 8 3 5 4 
 1 0    
 5 7  2 −1 5 12   3 1 2 − 2 

2. Find the value of k if the system equations of
+ − = 1, 2 + 3 + = 6 and 5 + 7 + = has solutions, and find these solutions.

3. Find the value of k such that the system equations of
+ + = 2, 2 − = 3 and + + 2 = has no solution.

4. Find the value of k such that the system equations of
+ + = 3, + 2 + 3 = 6 + 3 + = 4 + has infinite many solutions, and
find these solutions.

5. By using the Gaussian elimination, solve the following system of linear equations.
(a) x − 2y − 6z = 3 , x + 2z = y , 2x − 3y − z = 6

(b) 7x + 16 y − 10z = 9 , 5x + 11y − 8z = −6 , x + y = z

(c) y + 7z = 5x + 8 , x + 7 y = 5z − 16 , 7x + z = 5y + 14

(d) x − 2 y − z = −8 , 3x − y − 4z = −15 , y + 2z = 4

(e) x + 2y + 3z = 12 , 2x + 3y + z = 7 , 3x + y + 2z = 5

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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 2 −1 1 
6. It is given that P = 1 −1 −1 , find P −1 using elementary row operations.

 2 − 2 −1
Hence, solve the following system of linear equations.

2x − y + z = 12
x−y−1 z =3

2
x− y− z =1

7. A system of linear equations is given by
2x −3y + 4z =1
3x − y = 2
x + 2 y + kz = 1

where k is a real constant.
(a) Reduce the augmented matrix of the system of linear equations to row-echelon form .
(b) State the value of k such that the system has no unique solution. For this value of k,

solve the system of equations.

8. The variables x, y and z satisfy the system of linear equations
2x − 5y + 5z = −7
x − 2 y + 3z = −1
− x + 3y = 10

Write a matrix equation for the system of linear equations. By using Gaussian elimination, find
the variables x, y and z.

9. A system of linear equations is given by

2x + y + z = 2 , 4x − 2y + z = 7 , 6x − y + 2z = 9

Reduce the augmented matrix of the system of linear equations to row-echelon form.
Hence, find the solutions of the system.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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ANSWER 1(a) z = 1 , y = 2 , x = -3
Section System has unique solution.
3.2
1(b) System has no solution.
Exercise
3.5 1(c) Infinite many solution.

1(d) Infinite many solutions.

x = 4t, y = −3t, z = t

1(e) Unique solution.

z = 3 , y = −1 , x = −2.

1(f) Systems has no solutions.

2 System has solution

k = 13

x = 4t – 3 , y = 4 – 3t , z = t
3 System has no solution

k=3

4 System has infinite many solutions

k=5
x = t , y = 3 – 2t , z = t
5(a) z = 1 , y = −11 , x = −13.
5(b) x = 7 , y = 5 , z = 12

5(c) z = 2 , y = −1 , x = 1.

5(d) z = 1 , y = 2 , x = −3.
5(e) z = 3 , y = 2 , x = −1.

6  1 3 − 2
1 4 −3
P −1 =

0 − 2 1 

 x = 3 , y = −2 , z = 4

7(a)  2 − 3 4 1 
 0 − 7 12 − 1
 
 0 0 8 + 2k 0 

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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7(b) k = -4

∴ x = t, y = 3t – 2, z = 7t - 5
4

8 z = 2 , y = 3 , x = −1

9  2 1 1 2 
0 4 1 − 3
 
 0 0 0 0  REF

y = t , z = −4t − 3 , x = 3t + 5

2

CLONE STPM

1. Solve the following system of linear equations using Gaussian elimination.
− 2 + = 2

2 − 3 + 4 = 5

− + 3 + 2 = 1

2. Solve the following system of linear equations.
− + 2 = 5

3 + 2 + = 10

2 − 3 − 2 = −10

111 9 3 −9

3. It is given = (1 3 6) and = (−9 −6 15).

123 3 3 −6

a) Find XY.

b) Hence, find the values of a, b, c if

4
( ) = (3).

4

4. Solve the following system of linear equations using Gaussian elimination.
+ + = 6

2 + 3 + 7 = 29

+ 2 + 3 = 1

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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2 −1 1
5. It is given that P= (1 −1 −1).

2 −2 −1
a) Find P−1 using elementary row operations.
b) Hence, solve the following system of linear equations.

2 − + = 12
− − = 1

2 − 2 − = 6

123
6. a) Given that P= (2 4 5), find P−1 using elementary row operations.

356
b) Hence, solve the following system of linear equations.

+ 2 + 3 = −6
2 + 4 + 5 = −11
3 + 5 + 6 = −13

120 2 2 −2

7. Given A= (3 2 1) and B= ( 1 −1 1 ).

241 −8 0 4
a) Find AB.
b) Hence, determine −1.

c) The following table shows quantities of cupcakes and the amount paid by three customers.

Name of Strawberry Orange Chocolate Amount paid
customer cupcakes cupcakes cupcakes (RM)

Ali 1 2 0 5

Abu 3 2 1 10

Azman 2 4 1 13

The prices, in RM, of one strawberry cupcake, one orange cupcake and one chocolate
cupcake are x, y and z respectively.
(i) Write a system of linear equations for the above table.
(ii) Solve for the values of x, y and z in matrix form.

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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ANSWER 1 = −6, = −3, = 2.
Section
3.4 2 = 1, = 2, = 3

Clone 3(a) 3 0 0
STPM = (0 3 0)
003

3(b) = 3, = 2, = −1

4 = 1, = 2, = 3.
5(a)
1 3 −2
5(b) −1 = (1 4 −3)
6(a)
0 −2 1
6(b)
7(a) = 3, = −2, = 4

7(b) 1 −3 2
−1 = (−3 3 −1)
7(c) (i)
2 −1 0
7(c) (ii)
= 1, = −2, = −1

400
AB = (0 4 0)

004

0.50 0.50 −0.50
−1 = (0.25 −0.25 0.25 )

−2 0 1

+ 2 = 5

3 + 2 + = 10

2 + 4 + = 13

= 1, = 2, = 3

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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STPM PAST YEAR

STPM MT 2013 (Section A)

1. A system of linear equations is given by
+ + = ,
− + = 0,

4 + 2 + = 3,

where and k are real numbers. Show that the augmented matrix for the system may be
reduced to

1 1 1 [5 marks]
(0 −2 0 | − ).

0 0 − 4 3 − 3

Hence, determine the values of and k so that the system of linear equations has

(a) a unique solution, [ 1 mark ]
(b) infinitely many solutions, [ 1 mark ]
(c) no solution. [ 1 mark ]

STPM MT 2018 (Section A)

−3 2 −1

2. The matrix A is given by A = ( 1 1 1 )

201
By performing elementary row operations on the augmented matrix (A|I), where I is the 3 x 3

identity matrix, find A-1. [6 marks]

−3 2 −1 −7

Hence, solve the equation ( 1 1 1 ) ( ) = ( 2 ). [3 marks]

2 0 1 4

STPM MT 2019 (Section A)

3. A system of linear equations is given by

2 + 3 = −6,

4 − + 3 = 1,

2 − 4 + = 7,

where k is a real constant.

(a) Reduce the augmented matrix of the system of linear equations to row-echelon form.

[4 marks]

(b) State the value of k such that the system has infinitely many solutions, and find the general

form of the solutions. [5 marks]

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

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STPM MT 2015 (Section B)

4. The variables x, y and z satisfy the system of linear equations

2 + + 2 = 1,

4 + 2 + = ,

8 + 4 + 7 = 2,

where k is a real constant.

(a) Write a matrix equation for the system of linear equations. [1 mark]

(b) Reduce the augmented matrix to row-echelon form, and show that the system of linear

equations does not have a unique solution. [6 marks]

(c) Determine all the values of k for which the system of linear equations has infinitely many

solutions, and find the solutions in the case when k is positive. [6 marks]

(d) Find the set of values of k for which the system of linear equations is inconsistent.

[2 marks]

STPM MT 2017 (Section B)

5. A system of linear equations is given by
+ 2 + = ,
2 + + 4 = 3 ,
+ + = ,

where p and q are constants.

(a) Write the augmented matrix for the system above and reduce it to row-echelon form.

[5 marks]

(b) Determine the values of p and q such that the system has

(i) a unique solution, [3 marks]

(ii) no solution. [2 marks]

(c) Determine the values of p and q such that the system has infinitely many solutions. Using

the value of q and the smaller value of p obtained, find the solutions of the system.

[5 marks]

STPM MT 2020 (Section B)

2 − 1 2 2
6. Given a symmetric matrix, A = (2 − 1 ).

4 − 4 + 2 − 1

(a) Find the values of a, b and c. [4 marks]

(b) Using elementary row operations, determine the inverse of A for the values in (a).

[6 marks]

Hence, solve

(2 − 1) + 2 + 2 = 3

(2 − 1) + + = 11

(4 − 4) + ( + ) + (2 − 1) = 8 [5 marks]

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

212

ANSWER 1 a) ) ≠ 4
b) = 4 and = 1
Section c) = 4 and ≠ 1
3.4
2 x = 1, y = -1 and z = 2.
STPM
Past
Year

3(a) 2 3 0 −6
(0 7 −3 |−13)

0 0 − 3 0

3(b) k = 3

∴ The solution is 9 33 13
(− 14 − 14 , 7 − 7 , )

4(a) 2 1 2 1

(4 2 1) ( ) = ( )

8 4 7 2

4(b) Row 3 has all zero entries. So, the system of linear equations does not
have a unique solution.

4(c) ∴ = 2 since > 0
= 0
when z = 0; 2 + = 1
∴ If = , = 1 −

4(d) { : ≠ − 5 and ≠ 2}

3

5(a) 1 2 1

(0 1 1 − |0)

0 0 2 − 5 + 6

5(b) (i) For a unique solution,
≠ 2, ≠ 3, is any value

5(b) (ii) For no solution,
= 2 or = 3, ≠0

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

213

For infinitely many solution,
5(c) ∴ = 2, = 0 or = 3, = 0

The solutions of the system are (−3t , t , t ).

6(a) a = 1 , b = 2 , c = 2

6(b) 11
2 −2 0
1 7 4
−1 = −2 26 13

41
( 0 13 − 13)

= −4, = 51 and = 36 .
13 13

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG

214

UNIT SAINS DAN MATEMATIK, SEKTOR PEMBELAJARAN, JABATAN PENDIDIKAN NEGERI PULAU PINANG