Semester 1 Mathematics STPM (Teacher's Edition)

246

5 2 3 1 2 3 1 2 5 15
= (−1 0 4) (−1 0 4) = (2 −7 3 )
1 −1 1 1 −1 1 4 2 −2

2 3 1 2 5 15
= = (−1 0 4) (2 −7 3 )

1 −1 1 4 2 −2

14 −9 37
= (14 3 −23)

4 14 10

− 3 + 8 − 24

14 −9 37 2 5 15 2 31

= (14 3 −23) − 3 (2 −7 3 ) + 8 (−1 0 4)

4 14 10 4 2 −2 1 −1 1

100
− 24 (0 1 0)

001

000 [Shown]
= (0 0 0) =

000

− 3 + 8 − 24 =

[Multiply − on the right of both sides]

− − 3 − + 8 − − 24 − = − =

− 3 + 8 − 24 − =

24 − = − 3 + 8

− = 1 ( − 3 + 8 )
24

1 2 5 15 2 31 100

= 24 [(2 −7 3 ) − 3 (−1 0 4) + 8 (0 1 0)]
4 2 −2 1 −1 10 0 1

1 4 −4 12
= (5 1 −9)
24 1 5 3

1 11
6 −6 2

= 5 1 3
24 24 −8

15 1

(24 24 8 )

247

CLONE STPM

314
Given matrix = (1 1 4). Using elementary row operations, determine the inverse of A.
1. 4 4 3

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

(a) Show that 2 = , where I is the 3 × 3 identity matrix, and deduce −1.

1 43
(b) Find matrix which satisfies = ( 0 2 1).

−1 0 2

3. A, B and C are square matrices such that = − and = ( )− .

Show that − = = .

120
If = (0 −1 0), find C and A.

101

4. −1 2 1 −35 19 18

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

0 12 −3 12 5

Find the matrix and deduce the inverse of A.

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

156 −13 −50 −33 4 7 −13

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

1 −3 2 7 20 15 −2 1 11

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

248 1 3 1 41 0 0 →4 2− 3→ 3 3 1 4 10 0
(1 1 4|0 1 0) (1 1 4 |0 1 0)
ANSWER
Section 443 001 0 0 13 0 4 −1
3.1
3→ 2− 1→ 2 3 1 4 10 0
Clone (0 2 8 | −1 3 0)
STPM

0 0 13 0 4 −1

→2 1− 2→ 1 6 0 0 3 −3 0
(0 2 8 | −1 3 0)

0 0 13 0 4 −1

→13 2−8 3→ 2 6 0 0 3 −3 0
(0 26 0 | −13 7 8)

0 0 13 0 4 −1

1 10 0 1 − 1 0
6→ 1→ 1 0 26 2 2 8
0 | −13 7

(0 0 13 0 4 −1)

1 10 0 1 − 1 0
→26 2→ 2 0 1 2 2
|| 1 7 4
0 − 2 13
26
(0 0 13 0 4 −1)

1 100 1 − 1 0
2 2
1→3 3→ 3 0 | 1 7 4
1 0 | − 2 26 13

(0 0 1 0 4 − 1
13 13)

11
2 −2 0
1 7 4
∴ Inverse of , −1 = −2 26 13

41
( 0 13 − 13)

2(a) 1 0 0 1 0 0 1 0 0
2 = (1 −1 0) (1 −1 0) = (0 1 0) = (Shown)
1 −2 1 1 −2 1 0 0 1

−1 2 = −1 [Multiply A−1 on the left of both sides]

= −1

100
∴ −1 = = (1 −1 0)

1 −2 1

249

2 (b) 1 43
= ( 0 2 1)

−1 0 2

[Multiply A−1 on the right of both sides]

1 43 1 0 0
−1 = ( 0 2 1) (1 −1 0)

−1 0 2 1 −2 1

8 −10 3
∴ = (3 −4 1)

1 −4 2

3 = − [Multiply − at the right on both sides]
− = − − [Multiply B at the left on both sides]
= − −
= − −
= −
∴ − =

= ( )−

( )( ) = ( )( )− [Multiply AB at the left on both sides]

( ) =

− =

=

− = − [Multiply − at the left on both sides]

∴ = −

∴ − = = [Shown]

1 2 0 1 2 0 100
= = (0 −1 0) (0 −1 0) = (0 1 0)

1 0 1 1 0 1 221

1 0 01 0 0 → 3−2 1→ 3 1 0 01 00
( | ) = (0 1 0|0 1 0) (0 1 0| 0 1 0)

2 2 10 0 1 0 2 1 −2 0 1

→ 3−2 2→ 3 1 0 01 00
(0 1 0| 0 1 0)

0 0 1 −2 −2 1

= ( | − )

1 0 0 −1 1 00
= ( − )− = − = (0 1 0) = ( 0 1 0)

221 −2 −2 1

250

4 −1 2 1 −1 2 1 −35 19 18
= = (−3 1 4) (−3 1 4) (−27 −13 45)
0 1 2 0 1 2 −3 12 5

−1 2 1 −22 −33 77
= (−3 1 4) ( 66 −22 11)

0 1 2 −33 11 55

121 0 0
= ( 0 121 0 )

0 0 121

= 121

− = 121 −

− = 1 = 1 −1 2 1 −35 19 18
121 121 (−3 1 4) (−27 −13 45)
1 12
0 2 −3 5

23 7
− 11 − 11
1 −22 −33 77 6 2 11
= ( 66 −22 11) = 1
121 −33 11 11 − 11
55 3 1 11
5
(− 11 11 11)

5 1 5 6 −13 −50 −33 24 40 32

= (2 −2 4) ( −1 −6 −5 ) = ( 4 −8 4 )

1 −3 2 7 20 15 4 8 12

24 40 32 4 7 −13 72 0 0

= ( 4 −8 4 ) ( 1 −5 −1 ) = ( 0 72 0 )

4 8 12 −2 1 11 0 0 72

= 72

( )− = 72( )− [Multiply (PQ)−1 at the left of both sides]

= 72( )−

17 13
18 72 − 72
( )− = 1 = 1 4 7 −13 1 51
72 72 (1 −5 −1 ) = 72 − 72 − 72
1 1 1 11
−2 11

(− 36 72 72 )

251

STPM PAST YEAR

STPM MT 2014 (Section A)
−5 0 2

1. Matrix P is given by = ( 0 2 −1).

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

STPM MT 2016 (Section A)

1 12

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

(a) Find 2 − 6 + 11 . −1 1 3

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

ANSWER

Section 1. −5 0 2 1 0 0
( 0 2 −1 | 0 1 0)
3.1
−1 4 −2 0 0 1

STPM → 1−5 3→ 3 −5 0 2 10 0
Past Year (0 2 −1 | 0 1 0 )

0 −20 12 1 0 −5

→10 2+ 3→ 3 −5 0 21 0 0
(0 2 −1 | 0 1 0)

0 0 2 1 10 −5

→ 3− 1→ 1 5 0 00 10 −5
(0 2 −1 | 0 1 0)

0 0 2 1 10 −5

→0.5 3→ 3 5 0 00 10 −5
(0 2 −1 | 0 1 0)

0 0 1 0.5 5 −2.5

→ 2+ 3→ 2 5 0 00 10 −5
(0 2 0 | 0.5 6 −2.5)

0 0 1 0.5 5 −2.5

1 1 0 00 2 −1
(0 2 0 | 0.5 6 −2.5)
→5 1→ 1

0 0 1 0.5 5 −2.5

1 1 0 00 2 −1
(0 1 0 | 0.25 3 −1.25)
2→ 2→ 2

0 0 1 0.5 5 −2.5

0 2 −1
∴ −1 = (0.25 3 −1.25)

0.5 5 −2.5

252

2. (a) 2 − 6 + 11

1 12 1 12 1 12 100

= ( 0 2 2) ( 0 2 2) − 6 ( 0 2 2) + 11 (0 1 0)

−1 1 3 −1 1 3 −1 1 3 001

−1 5 10 6 6 12 11 0 0

= (−2 6 10) − ( 0 12 12) + ( 0 11 0 )

−4 4 9 −6 6 18 0 0 11

4 −1 −2
= (−2 5 −2)

2 −2 2

2. (b) 1 1 2 4 −1 −2
( 2 − 6 + 11 ) = ( 0 2 2) (−2 5 −2)

−1 1 3 2 −2 2

600
= (0 6 0) = 6 (Shown)

006

[Multiply −1 on the left of both sides]

−1 ( 2 − 6 + 11 ) = 6 −1

( 2 − 6 + 11 ) = 6 −1

∴ −1 = 1 ( 2 − 6 + 11 ) = 1 4 −1 −2
6 6 (−2 5 −2)
−2
2 2

2 11
3 −6 −3

= 1 5 1
−3 6 −3
1 11
( 3 −3 3 )

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

Quick Notes 3.4

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

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

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

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

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

Coincident lines
x+3y=1

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

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

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

11 12 13 1

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

For example, the equations

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

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

254

Finding the unique solution using the inverse matrix

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

A-1.

AX = B (given system)

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

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

IX = A-1B (inverse property)

X = A-1B (identity property)

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

Solving system of Linear Equations (Gaussian Elimination Method)

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

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

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

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

Gaussian elimination Method

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

unknowns:

x+ y+ z=3

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

x + 3y + 2z = 3

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

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

x + 3y + 2z = 3 1 3 2 3

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

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

1 3 2 3   1 3 2 3 
  

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

255

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

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

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

2 y + 4z = −6

− 3z = 6

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

Difference between Gaussian Elimination and Gaussian Jordan Elimination

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

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

0 0 f 

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

0 0 1

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

Solve the following system of linear equations.

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

Solution:

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

To find inverse of matrix A using elementary row operation.

256

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

0 −1 8 0 0 1 0 −1 8 0 0 1

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

0 0 1 −1 1 1

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

0 0 1 −1 1 1

24 −23 −21
A−1 = (−8 8
7)

−1 1 1

Therefore, X = A-1B
24 −23 −21

= (−8 8 7 ) ( )

−1 1 1

−85
= ( 29 )

4

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

EXAMPLE 2: Finding Inverse By Elementary Row Operations

1 −2 1
If P = ( 3 1 2) , find P-1.

−1 4 1

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

3x + y + 2z = 4

-x + 4y + z = 2

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

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

−1 4 1 0 0 1

2 − 3 1 → 2

3 + 1 → 3

257

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

7 77

0 2 2 1 01

1 + 2 2 → 1 10 5 1 20
3 − 2 2 → 3
77 7

0 1 − 17||− 3 1 0
7 7

(0 0 16 13 −2 1)
77
7

10 5 1 20

77 7

7 3 → 3 0 1 − 17||− 3 1 0
16 7 7

(0 0 16 13 −2 1)
77
7

5 −7 6 −5
1 − 7 3 → 1 1 0 0 16 16 16

1 0 1 0||− 5 2 1
2 + 7 3 → 2 16 16 16
0 0 1 13
( 16 −2 7

16 16 )

Hence, P-1 = 7 6 5
− 16 − 16
16
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
−7 6 −5 1
1
= 16 (−5 2 1 ) (4)
13 −2 72

258

7

16

=5

16
19

75 (16)
16 , = 16 19
Hence, = = 16

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:

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

5x + 11y − 8z = −6

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

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

259

⎯2⎯R2−⎯13R⎯3→⎯R3 →  2 3 4 1 
 0 13 16 − 3 
 − 33 − 33 REF
 0 0

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

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

2 − 2 1 → 2 1 1 −1 −2
1 + 3 → 3 (0 −3 3 | 9 )

0 3 1 −1

2 + 3 → 3 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.

260

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:

2 1 −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,

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

0 5 7 1 13

1 1 −1 1
(0 1 3 |4)
2 68
0
1 −1 1
1 3 |4)
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.

261

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

1↔ 2 1 −1 −2 2
(3 −4 4 |7)

2 −3 6 5

2 − 3 1 → 2 1 −1 −2 2
3 − 2 1 → 3 (0 −1 10|1)

0 −1 10 1

2 − 3 → 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.

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

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

x-y=5

262

Solution :

The augmented matrix for this system is
2 −2 −1 2

(4 −4 −3|2)
1 −1 0 5

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

R3-R2 → R3 0 0 −1 −8

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 numbers. 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
4k 7k 74

2(2k +12)−1(3k + 21)−1(12 −14) = 0

k = −5

263

 2 1 − 1 2  1 1 −2 0  3R1 − R2 → R2  1 1 − 2 0 
3 2 − 3 2 R2 − R1 → R1  3 2 −3 2 7 R1 − R3 → R3 0 1 −3 − 2 
 4 − 5 6  7 4 −5   3 − 9 − 6
 7 6   0

⎯⎯⎯⎯⎯→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:

a) 2x − 2y + z =1   2 −2 1 1 
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⎯)R2−⎯7R3⎯→⎯R3 →  2 −2 1 1 
0 7 −5 4k − 2 

 0 0 5− 5k 4k 2 + 22k − 26  REF
 

b) For unique solution: 5 − 5k  0  k  1 z=−2
For k = −6 , 5

Row 3 → 0x + 0y + 35z = −14  y = −4

Row 2 → 0x + 7 y − 5z = −26

Row 1 → 2x − 2y + z = 1  x = − 33

10

264

5 − 5k = 0 4k 2 + 22k − 26 = 0
c) For infinitely many solutions: k = 1 and
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: 2 1 12
Using Gaussian elimination, (8 3 5| 4 )

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

6 2 4 −4

-3R3 + R3 → R3 2 1 12
(0 −1 1| −4 )

0 −1 1 −10

-R2 + R3 → 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.

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.

265

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 1 2 4   6 −3 0 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 
     2 − 2
 5 7 1 0   2 −1 5 12   3 1

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 − 2 y − 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 + 2 y + 3z = 12 , 2x + 3y + z = 7 , 3x + y + 2z = 5

266

 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 torow-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.

267

ANSWER 1(a) 1 −2 −1 −8
Section (3 −1 −4|−15)
3.2 01 24

Exercise 2 − 3 1 → 2 1 −2 −1 −8
3.5 (0 5 −1| 9 )

0 1 24

5 3 − 2 → 3 1 −2 −1 −8
(0 5 −1 | 9 )

0 0 −11 11

∴ 11z = 11 ⟹ z = 1
5y – z = 9 ⟹ y = 2
x - 2y – z = - 8 ⟹ x = -3

System has unique solution.

1(b) 4 −2 4 2
(2 −1 −4|4)
6 −3 0 3

1 + 2 − 3 → 3 4 −2 4 2
(2 −1 −4|4)

0 0 03

Since 0x + 0y + 0z = 3 (inconsistent)
∴ System has no solution.

1(c) 1 2 3 3
(2 3 8 |4)
3 2 17 1

2 1 − 2 → 2 1 2 33
3 1 − 3 → 3 (0 1 −2|2)

0 4 −8 8

4 2 − 3 → 3 1 2 33
(0 1 −2|2)

0 0 00

Let z = t : y – 2z = 2 ⟹ y = 2 + 2t
x + 2y + 3z = 3 ⟹ x = -1 -7t
∴ x = -1 – 7t , y = 2 + 2t , z = t
∴ Infinite many solution.

268

1(d) 1 1 − 1 0 1 1 − 1 0
0 1 3 0
2 3 1 0 R2 − 2R1 → R2 0 2 6 0
  R3 − 5R1 → R3
 5 7 1 0 

R3 − 2R2 → R3

1 1 − 1 0
0 1 3 0
 0 0
 0 0

Infinite many solutions.

Let z = t : y + 3z = 0  y = −3t

x + y − z = 0  x = 4t

 x = 4t, y = −3t, z = t

1(e)  4 − 4 3 5 

1 − 2 1 3  R3 − 2R2 → R2
  2R3 − R1 → R3
 2 −1 5 12 

 4 − 4 3 5 
0 3 3 6 
 
 0 2 7 19 

R3 − 2 R2 → R3  4 − 4 3 5 
3 0 3 3 6 

 0 0 5 15 
 

1 R2 → R2
3

 4 − 4 3 5 
0 1 1 2 
 5 15
 0 0

Unique solution.

5z = 15  z = 3.
y + z = 2  y = −1
4x − 4 y + 3z = 5  x = −2.

269

1(f)  2 1 1 2  R1 − R2 + 2R3 → R3
8 3 5 4 
 
 3 1 2 − 2 

 2 1 1 2 
8 3 5 4 
 0 − 6
 0 0

Since 0x + 0 y + 0z = −6 (inconsistent)

Systems has no solutions.

2 1 1 −1 1
(2 3 1 |6)
5 7 1

1 + 2 2 − 3 → 3 1 1 −1 1
(0 1 3 | 4 )
2 − 2 1 → 2
0 0 0 13 −
Let z = t : y +3z = 4 ⟹ y = 4 – 3t
x + y – z = 1 ⟹ x = 4t – 3 System has solution : 13 – k =
∴ x = 4t – 3 , y = 4 – 3t , z = t 0 ⟹ k = 13

3 1 1 2
(2 0 −1|3)
1 1 2

3 + 2 − 1 → 3 1 1 2
( 2 0 −1| 3 )

3 − 0 0 + 1

System has no solution : 3 – k = 0 and k + 1 ≠ 0
∴k=3

4 111 3
(1 2 3| 6 )
1 3 4 +

1 + 3 − 2 2 → 3 11 1 3
(0 1 2 | 3 )
2 − 1 → 2
0 0 − 5 − 5
Let z = t : y +2z = 3 ⟹ y = 3 – 2t
x+y+z=3⟹x=t System has infinite many
∴ x = t , y = 3 – 2t , z = t solutions : k - 5 = 0 ⟹ k = 5

270

5(a) x − 2y − 6z = 3   1 −2 − 6 3
x + 2z = y  1 −1 2 0

  − 1 6
2x − 3y − z = 6  2 −3

R2 − R1 → R2 1 − 2 − 6 3 
0 1 8 − 3
R3 − 2R1 → R3  0 1 
11 0 

R3 − R2 → R3 1 − 2 − 6 3 

0 1 8 − 3
 
 0 0 3 3 

3z = 3  z = 1.
y + 8z = −3  y = −11
x − 2 y − 6z = 3  x = −13.

5(b) 7x + 16 y − 10z = 9  7 16 − 10 9 
5x + 11y − 8z = −6   5 11 − 8 − 6

  
x+ y=z  1 1 −1 0 

10R3 − R1 → R1  3 −6 0 9
8R3 − R2 → R2 3 −3 0 6
1 1 
−1 0 

2R2 − R1 → R1 3 0 0 21

3 − 3 0 6 
1 
1 −1 0 

3x = 21  x = 7.
3x − 3y + 0z = 6  y = 5
x + y − z = 0  z = 12.

271

5(c) y + 7z = 5x + 8  x + 7 y − 5z = −16   1 7 − 5 − 16
x + 7 y = 5z − 16  − 5x + y + 7z = 8  − 5 1 7 8
 
7x + z = 5y + 14    
7x − 5y + z = 14  7 −5 1 14 

 1 7 − 5 − 16 5R1 + R2 → R2  1 7 − 5 − 16 
− 5 1 7 8 7R1 − R3 → R3 0 36 − 18 − 72 
 −5   54 
 7 1 14   0 − 36 − 126 

1 R2 − 1 R3 → R3  1 7 − 5 − 16 
6 9 0 36 − 18 − 72

 0 0 1 2 
 

1 R2 → R2  1 7 − 5 − 16
18 0 2 −1 − 4 

 0 0 1 2 
 

z = 2.
2 y − z = −4  y = −1
x + 7 y − 5z = −16  x = 1.

5(d) x − 2y − z = −8   1 −2 − 1 − 8 
3x − y − 4z = −15  3 −1 − 4 − 15

  
y + 2z = 4  0 1 2 4 

R2 − 3R1 → R2 1 − 2 −1 − 8
0 5 −1 9 
 
 0 1 2 4 

5R3 − R2 → R3  1 −2 − 1 − 8
0 5 −1 9 

 0 0 11 11 
 

11z = 11  z = 1.
5y − z = 9 y = 2
x − 2 y − z = −8  x = −3.

272

5(e) x + 2 y + 3z = 12   1 2 3 12
2x + 3y + z = 7  2 3 1 7

  
3x + y + 2z = 5  3 1 2 5 

2R1 − R2 → R2 1 2 3 12
3R1 − R3 → R3 0 1 5 17 
 7 31
 0 5

5R2 − R3 → R3 1 2 3 12 
 0 1 5 17 
 
 0 0 18 54 

18z = 54  z = 3.

y + 5z = 17  y = 2

x + 2 y + 3z = 12  x = −1.

6  2 −1 1 1 0 0
1 −1 −1 0 1 0
−2 −1 0 0 1  R1  R2

 2


 1 −1 −1 0 1 0
 2 −1 1 1 0 0
 1 
 2 −2 −1 0 0

− 2R1 + R2 → R3  1 −1 −1 0 1 0 
2R1 + R3 → R2  0 1 31 −2 0
 0 0 10 −2 1 


R2 + R1 → R1  1 0 2 1 −1 0
 0 1 3 1 − 2 0
 
 0 0 10 −2 1 

− 3R3 + R2 → R2  1 0 0 1 3 − 2
0 1 01 4 − 3
− 2R3 + R1 → R1  0 10 −2 
 0 1 

P −1 =  1 3 − 2
 1 4 −3

0 − 2 1 

2x − y + z = 12 2x − y + z = 12

x− y− 1 z =3 → x− y− z =1
2
x− y− z =1 2x − 2y − z = 6

273

 2 −1 1  x  12
1 −1 −1 y  =  1 
 2 − 2 −1 z   6 

 x  1 3 − 212  3 
 y = 1 4 − 3 1  =  − 2
 z  0 − 2 1  6   4 

 x = 3 , y = −2 , z = 4

7(a)  2 −3 4 1
−1
3 2 0 2 3R1 − 2R2 → R2
  R1 − 2R3 → R3
 1 k 1 

 2 − 3 4 1 
 0 − 7 12 − 1
 4 − 2k − 1
 0 −7

R2 − R3 → R3

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

7(b) 8 + 2k = 0
k = -4

2x – 3y + 4z = 1 ⇒ 6x – 9y + 12z = 3 --- (1)

-7y + 12z = -1 ---(2)
(1) – (2), 6x – 2y = 4

y = 3x – 2

let x = t, y = 3t – 2
-7(3t – 2) + 12z = -1
z = 7t - 5

4

∴ x = t, y = 3t – 2, z = 7t - 5

4

274

8  2 −5 5 x   −7 
     −1
 1 −2 3   y  = 

 −1 3 0 z   10 

 2 − 5 5 − 7
 1 − 2 3 −1
 
 − 1 3 0 10 

⎯⎯R1⎯R2 →  1 − 2 3 − 1
 2 − 5 5 − 7
 
 − 1 3 0 10 

− 2R1 + R2 → R2 1 − 2 3 − 1
0 −1 −1 − 5
R1 − R3 → R3  
 0 1 3 9 

− R2 → R2 1 − 2 3 − 1
R2 + R3 → R3 0 1 1 5 
 
 0 0 2 4 

Solve Row 3 → 0x + 0y + 2z = 4  z = 2

Solve row 2 → 0x + 1y + (2) = 5  y = 3

Solve row 1 → x − 2(3) + 3(2) = −1  x = −1

9 2x + y + z = 2   2 1 1 2
4x − 2y + z = 7  4 −2 1 7

  
6x − y + 2z = 9  6 −1 2 9 

2 R1 − R2 → R2  2 1 1 2 
3R1 − R3 → R3 0 4 1 − 3
 4 1 − 3
 0

⎯⎯R3−R⎯2→⎯R3 →  2 1 1 2 
0 4 1 − 3
Let y = t  
 0 0 0 0  REF

Solve row 2 → 0x + 4y + +z = −3  z = −4t − 3

Solve row 1 → 2x + y + z = 2  x = 3t + 5

2

275

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

276

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.

ANSWER 1. 1 −2 1 2
Section ( 2 −3 4|5)
3.4 −2 1 + 2 → 2
1 + 3 → 3 −1 3 2 1
Clone
STPM − 2 + 3 → 3 1 −2 1 2
2 2 + 1 → 1 (0 1 2|1)
−2 3 + 2 → 2
−5 3 + 1 → 1 0 1 33

1 0 54
(0 1 2|1)

0 0 12

1 0 54
(0 1 0|−3)

0 0 01

∴ = −6, = −3, = 2.

2 1 −1 2 5
(3 2 1 | 10 )
2 −3 −2 −10

→ 2−3 1→ 2 1 −1 25
(0 5 −5| −5 )
3−2 1→ 3 0 −1 −6 −20

1 1 −1 25
(0 1 −1| −1 )
5→ 2→ 2

0 −1 −6 −20

→ 1+ 2→ 1 1 0 14
(0 1 −1| −1 )
2+ 3→ 3 0 0 −7 −21

277

1 1 0 14
(0 1 −1|−1)
−→7 3→ 3

00 13

→ 1− 3→ 1 1 0 01
(0 1 0|2)
2+ 3→ 2 0 0 1 3

∴ = 1, = 2, = 3

3(a) 1 1 1 9 3 −9
= (1 3 6) (−9 −6 15)
1 2 3 3 3 −6

300
= (0 3 0)

003

3(b) 1 0 0
= 3 (0 1 0)
001

1 100
(3 ) = (0 1 0)
0 0 1

−1 = 1 = 1 9 3 −9
3 3 (−9 −6 15 )
3 −6
3

3 1 −3
= (−3 −2 5 )

1 1 −2

4
( ) = (3)

4

3 1 −3 4
( ) = (−3 −2 5 ) (3)

1 1 −2 4

3
=(2 )

−1

∴ = 3, = 2, = −1

278 11 16
4 (2 3 7 |29)

2−2 1 → 2 1 3 −2 1
3 − 1 → 3
11 16
(0 1 5 |17)

0 2 −3 −5

3 − 2 2 → 3 11 1 6
(0 1 5 | 17 )

0 0 −13 −39

1 1 1 16
− 13 3 → 3 (0 1 5|17)

0 0 13

2 − 5 3 → 2 1 1 16
(0 1 0|2)

0 0 13

1 − 3 → 3 1 0 14
(0 1 0|2)

0 0 13

1 − 3 → 3 1 0 01
∴ = 1, = 2, = 3. (0 1 0|2)

0 0 13

279
5(a)

2 −1 1 1 0 0
(1 −1 −1|0 1 0)

2 −2 −1 0 0 1

1 ↔ 2 1 −1 −1 0 1 0
(2 −1 1 |1 0 0)

2 −2 −1 0 0 1

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

0 0 1 0 −2 1

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

0 0 1 0 −2 1

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

0 −2 1

280

5(b) 2 −1 1 12
(1 −1 −1) ( ) = ( 1 )

2 −2 1 6

1 3 −2 12
( ) = (1 4 −3) ( 1 )

0 −2 1 6

3
= (−2)

4

∴ = 3, = −2, = 4

6(a) 1 2 3 1 0 0
(2 4 5|0 1 0)
3 5 60 0 1

−→3 1+ 3 → 3 1 2 31 00
(0 0 −1|−2 1 0)
−2 1+ 2→ 2 0 −1 −3 −3 0 1

→ 2↔ 3 1 2 31 0 0
(0 −1 −3|−3 0 1)

0 0 −1 −2 1 0

−→ 2→ 2 1 2 31 0 0
(0 1 3|3 0 −1)
− 3→ 3 0 0 1 2 −1 0

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

→3 3+ 1→ 1 1 0 01 −3 2
(0 1 0|−3 3 −1)

0 0 1 2 −1 0

1 −3 2
∴ −1 = (−3 3 −1)

2 −1 0

281

6(b) 1 2 3 −6

(2 4 5) ( ) = (−11)

3 5 6 −13

1 −3 2 −6
( ) = (−3 3 −1) (−11)

2 −1 0 −13

1
= (−2)

−1

∴ = 1, = −2, = −1

7(a) 1 2 0 2 2 −2
AB=(3 2 1) ( 1 −1 1 )
2 4 1 −8 0 4

400
= (0 4 0)

004

7(b) AB = 4I

A (14 B) = I
∴ −1 = 1
4
2 2 −2
−1 = 1 (1 −1 1)
4 0 4
−8

0.50 0.50 −0.50

= (0.25 −0.25 0.25 )

−2 0 1

7(c) (i) + 2 = 5
3 + 2 + = 10
2 + 4 + = 13

7(c) (ii) 1 2 0 5

(3 2 1) ( ) = (10)

2 4 1 13

0.50 0.50 −0.50 5

( ) = (0.25 −0.25 0.25 ) (10)

−2 0 1 13

1
= (2)

3

∴ = 1, = 2, = 3

282
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]

283

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]

284

ANSWER 1 1 1 1
(1 −1 1|0)
Section 4 2 0
3.4
2 − 1 → 2 1 1 1
STPM (0 −2 0|− )
Past
Year 4 2 3

3 − 4 1 → 3 1 1 1
(0 −2 0 | − )

0 −2 − 4 3 − 4

3 − 2 → 3 1 1 1
(0 −2 0 | − )

0 0 − 4 3 − 3

a) ) ≠ 4
b) = 4 and = 1
c) = 4 and ≠ 1

−3 2 −1 1 0 0
2 ( 1 1 1 |0 1 0)

2 0 10 0 1

1 + 3 → 1 1 2 01 0 1
2 − 3 → 2 (−1 1 0|0 1 −1)

2 0 10 0 1

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

2 0 10 0 1

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

0 0 1 −2 4 −5

1 −2 −3
Hence, −1 = ( 1 −1 2 )

−2 4 −5

−3 2 −1 −7

( 1 1 1 ) ( ) = ( 2 )

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

−2 4
1

( ) = (−1)

2

Hence, x = 1, y = -1 and z = 2.

285

3(a)
2 3 0 −6

(4 −1 3| 1 )
2 −4 7

2 1 − 2 → 2 2 3 0 −6
1 − 3 → 3 (0 7 −3|−13)
2 − 3 → 3
0 7 − −13

2 3 0 −6
(0 7 −3 |−13)

0 0 − 3 0

3(b) For infinite many solutions,
k–3=0

k=3

For general form of the solutions,

Let z = t : 7y – 3t = -13 ⟹ y = 3 − 13
7 7
2 + 3 (37 − 173) = −6
2 = − 9 − 3

77

= − 9 − 3

14 14

∴ The solution is 9 − 3 3 − 13
(− 14 14 , 7 7 , )

4(a) 2 1 2 1

(4 2 1) ( ) = ( )

8 4 7 2

2 1 21
4(b) (4 2 1| )

8 4 7 2

2 − 2 1 → 2 21 2 1
3 − 4 1 → 3
(0 0 −3| − 2 )
0 0 −1 2 − 4

2 − 3 3 → 2 21 2 1

(0 0 0 |−3 2 + + 10)

0 0 −1 2 − 4

3 ↔ 2 212 1

(0 0 1| 4 − 2 )

0 0 0 −3 2 + + 10

Row 3 has all zero entries. So, the system of linear equations does not

have a unique solution.

286

4(c) The system of linear equations has infinitely many solutions when
3 2 − − 10 = 0

(3 + 5)( − 2) = 0

= − 5 or = 2

3

∴ = 2 since > 0

21 2 1 2 1 21

when k = 2; (0 0 0 |−3 2 + + 10) → (0 0 0 |0)

0 0 −1 2 − 4 0 0 −1 0

⇒ 2 + + 2 = 1 and − = 0 ⇒ = 0

when z = 0; 2 + = 1
∴ If =

= 1 −

4(d) The system of linear equations is inconsistent.
when 3 2 − − 10 ≠ 0

(3 + 5)( − 2) ≠ 0

or { : ≠ −5 ≠ 2}

3

1 2 1
5(a) (2 4|3 )

1 1

1 − 3 → 2 12 1
2 1 − 2 → 3
(4 − ) 2 − 3 → 3 (0 1 1 − | 0 )

0 4 − −2 −

12 1

(0 1 1 − |0)

0 0 (4 − )(1 − ) + 2

12 1

(0 1 1 − |0)

0 0 2 − 5 + 6

5(b) (i) For a unique solution,

2 − 5 + 6 ≠ 0

( − 2)( − 3) ≠ 0
≠ 2, ≠ 3, is any value

287

5(b) (ii) For no solution,
2 − 5 + 6 = 0, ≠ 0
( − 2)( − 3) = 0
= 2 = 3, ≠0

For infinitely many solution,
5(c) 2 − 5 + 6 = 0, = 0

( − 2)( − 3) = 0
= 2 = 3

∴ = 2, = 0 or = 3, = 0

= 2, = 0

1 2 10
(0 1 −1|0)

0 0 00

x + 2y + z =0
y–z=0

Let z = t

y=t
x = - 2t – t = -3t

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

6(a) 2 − 1 2 2
= (2 − 1 )

4 − 4 + 2 − 1

A is a symmetric matrix.

2 = 2 − 1 ………….○1
2 = 4 − 4 ………….○2
= + …………….○3
From ○1 , 2 − 2 + 1 = 0

( − 1)2 = 0
a=1

From ○2 , 2 − 4 + 4 = 0

( − 2)2 = 0
b=2

substitute b = 2 into ○3 ,

2c = 2 + c

c=2

288

314
6(b) = (1 1 4)

443 3 1 41 0 0

(1 1 4|0 1 0)

4 4 30 0 1

1 ↔ 2 1 1 40 1 0
(3 1 4|1 0 0)

4 4 30 0 1

2 − 3 1 → 2 1 1 40 1 0
3 − 4 1 → 3 (0 −2 −8 |1 −3 0)

0 0 −13 0 −4 1

1 1 1 44|−021 1 0
− 2 2 → 2 (0 1 0)
3
1 2
− 13 3 → 3
0 0 1 0 4 −1

13 13

2 − 4 3 → 2 0 − 3 4
1 − 4 3 → 1 110 13 13

0 1 0||− 1 7 4
2 26 13
0 0 10
( 4 − 1
13
13)

1 − 1 0
2
1 0 02
1 − 2 → 1 0||− 1 7 4
0 1 2 26 13

0 0 10 4 − 1
( 13
13)

11
2 −2 0
1 7 4
Hence, −1 = −2 26 13

41
( 0 13 − 13)

2 − 1 2 2 3

(2 − 1 ) ( ) = (11)

4 − 4 + 2 − 1 8

3 1 4 3

(1 1 4) ( ) = (11)

4 4 3 8

289

3
( ) = (11)

8

3
( ) = −1 (11)

8

1 − 1 0 3
2 2

= − 1 7 4 (11)
2 26 13
8
(0 4 − 1
13
13)

3 − 11 0

2 2
77
= −3 32
26
2 44 13

(0 13 − 8

13)

−4

51

= ( 13 )

36

13

Hence, = −4, = 51 = 36 .
13 13

290