maths

1. 1. Given
P =










112
121
211
(a) Find R so that R = P2
– 4P – I, [3 marks]
(b) Show that PR + 4I = 0. [2 marks]
Answer:
1.

2. 2. During the school’s family day, U6A offers three sales packages A, B and C for spaghetti, hot-dogs and cupcakes.
The number of each item and the offer price for each package are shown in the following table.
Sales Package Number of plates of
spaghetti
Number of hot-
dogs
Number of
cupcakes
Offer price (RM)
A 4 2 2 28
B 3 4 3 30
C 2 4 2 24
By representing the prices of a plate of spaghetti, a hotdog and a cupcake as x, y and z respectively, obtain the
simultaneous equation representing the information above. [3 marks]
By applying the Gaussian elimination, find the price of a plate of spaghetti, a hotdog and a cupcake respectively. [12
marks]
Answer :

5. The matrices A and B are given by 𝐴 = (
2 3 4
4 3 1
1 2 4
), 𝐵 = (
−10 4 9
15 −4 −14
−5 1 6
). State with reason
whether matrix 𝐴 is singular. Find the matrix 𝐴𝐵, and hence, deduce 𝐴−1
. [5 mark]
3. 𝐴 = (
2 3 4
4 3 1
1 2 4
) and | 𝐴| = [2 × 3 × 4 + 3 × 1 × 1 + 4 × 2 × 4] − [1 × 3 × 4 + 2 × 1 × 2 +
4 × 3 × 4] = −27 which is not equal to 0, 𝐴−1
is exist and is a singular.###
𝐴𝐵 = (
2 3 4
4 3 1
1 2 4
)(
−10 4 9
15 −4 −14
−5 1 6
)
= (
−20 + 3 × 15 − 20 2 × 4 − 12 + 4 × 1 18− 3 × 14+ 4 × 6
−40 + 45 − 5 4 × 4 − 3 × 4 + 1 36− 3 × 14+ 6
−10+ 2 × 15 − 4 × 5 4 − 2 × 4 + 4 9 − 2 × 14 + 4 × 6
)
= (
5 0 0
0 5 0
0 0 5
)
#####
= 5𝐼
𝐴𝐵 = 5𝐼
𝐴
𝐵
5
= 𝐼 => 𝐴𝐴−1
= 𝐼 =>
𝐵
5
= 𝐴−1
𝐴−1
=
1
5
(
−10 4 9
15 −4 −14
−5 1 6
)
###

6. Given that 𝐴 = (
1 3 1
1 −1 0
2 1 1
), show that 𝐴3
= 𝐴2
+ 6𝐴 − 𝐼, where I is the identity matrix. Hence, deduce 𝐴−1
.
[7 marks]
Answer:
𝐴3
= (
1 3 1
1 −1 0
2 1 1
) (
1 3 1
1 −1 0
2 1 1
)(
1 3 1
1 −1 0
2 1 1
) [M1]
= (
11 19 8
6 −3 1
17 12 8
) [A1]
𝐴2
+ 6𝐴 − 𝐼 = (
1 3 1
1 −1 0
2 1 1
) (
1 3 1
1 −1 0
2 1 1
) + 6(
1 3 1
1 −1 0
2 1 1
) − (
1 0 0
0 1 0
0 0 1
) [M1]
= (
11 19 8
6 −3 1
17 12 8
)
∴ 𝐴3
= 𝐴2
+ 6𝐴 − 𝐼 [A1]
𝐴3
𝐴−1
= 𝐴2
𝐴−1
+ 6𝐴𝐴−1
− 𝐼𝐴−1
[M1]
𝐴−1
= 𝐴 + 6𝐼 − 𝐴2
= (
1 3 1
1 −1 0
2 1 1
) + 6 (
1 0 0
0 1 0
0 0 1
) − (
1 3 1
1 −1 0
2 1 1
)(
1 3 1
1 −1 0
2 1 1
) [M1]
=(
1 2 −1
1 1 −1
−3 −5 4
) [A1]
[7 marks]

7. Matrix M is given by M =












01
103
101
k
.
Find M1 in terms of k using the elementary row operations. [5 marks]
Hence, solve the simultaneous equations
x – z + 3 = 0
3x + z – 11 = 0
4x + y – 5 = 0 [5 marks]
Answer :












100
010
001
01
103
101
k
B1
31
213
101
013
001
10
400
101
RR
RR
k 















M1
3
2
1
4
1
1
0
1
0
4
1
4
3
001
1
10
100
101
R
k
R
kk
k 



















32
21
1
1
4
1
4
1
0
4
1
4
3
0
4
1
4
1
010
100
001
RR
k
RR
kkk



















32
0
4
1
4
3
1
4
1
4
1
0
4
1
4
1
100
010
001
RRkkk 


















M1A1

















 
0
4
1
4
3
1
4
1
4
1
0
4
1
4
1
M 1
kkk
B1
x + 0y – z = –3
3x + 0y + z = 11
x – ¼ y + 0z = 5/4  k = ¼ B1



























 
013
1644
011
4
1
0
4
1
4
3
411
0
4
1
4
1
M 1







































4
5
11
3
0
4
1
1
103
101
z
y
x
B1





































4
5
11
3
013
1644
011
4
1
z
y
x
M1






















5
3
2
z
y
x
A1
 x = 2, y = 3, z = 5A1

8. Matrix P is given by
P =












131
021
001
for P2 + mP + nI = O.
(a) Find the values of m and n. [5 marks]
(b) Hence deduce P3 and P1. [5 marks]
Using an augmented matrix and elementary row operations, find the solution of the system of
equations.
422
253
62



zyx
zyx
zyx
(9 marks)

9. A pharmacist prescribed to a client 11 mg of vitamin A, 18 mg of vitamin B and 14 mg of vitamin C each day. The client
can choose from the combination of three types of tablets namely P,Q and R. Each tablet of P contains 1 mg of vitamin A, 1
mg of vitamin B and 2mg of vitamin C. Each tablet of Q contains 1 mg of vitamin A, 2 mg of vitamin B and 1 mg of vitamin
C. Each tablet of R contains 3 mg of vitamin A, 5 mg of vitamin B and 3 mg of vitamin C. The above information is
tabulated as follows.
Vitamin Type of tablets Amount of vitamin (mg)
P Q R
A 1 1 3 11
B 1 2 5 18
C 2 1 3 14
By using x as the number of tablets of P, y the number of tablets of Q and z the
number of tablets of R,
(a) form a system of linear equations based on the above information. [2 marks]
(b) use the Gauss-Jordan elimination method to solve the system of linear equations
described in (a).
(a) (i)
𝑥 + 𝑦 + 3𝑧 = 11
𝑥 + 2𝑦 + 𝑧 = 18
2𝑥 + 𝑦 + 3𝑧 = 14
(ii) (
1 1 3
1 2 5
2 1 3
) (
𝑥
𝑦
𝑧
) = (
11
18
14
)
(
1 1 3
1 2 5
2 1 3
|
11
18
14
)
−𝑅1+𝑅2
→ (
1 1 3
0 1 2
2 1 3
|
11
7
14
)
−2𝑅1+𝑅3
→ (
1 1 3
0 1 2
0 −1 −3
|
11
7
−8
)
−𝑅2+𝑅1
→ (
1 0 1
0 1 2
0 −1 −3
|
4
7
−8
)
𝑅2+𝑅3
→ (
1 0 1
0 1 2
0 0 −1
|
4
7
−1
)
−𝑅3
→ (
1 0 1
0 1 2
0 0 1
|
4
7
1
)
−𝑅3++𝑅1
→ (
1 0 0
0 1 2
0 0 1
|
3
7
1
)
−2𝑅3++𝑅2
→ (
1 0 0
0 1 0
0 0 1
|
3
5
1
)
x = 3, y = 5, z = 1
B1B1
M1 M1
M1M1M1
M1M1
A1