Matrices: Mathematics Tingkatan 5 questions

Section A: objective questions

11 marks

Choose the correct answer. Each question carries 1 mark.

Which of the following is a column matrix?

2−58
2−581
2−58
2−58103

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A column matrix has only one column. The first option is a row matrix, the second is square and the last is 3 × 2.

For the teacher: Remembering, low difficulty, standard 2.1.1

Textbook p. 38

Given A = 7−2509−4, what is a₂₃?

5
−4
9
0

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a₂₃ is the element in row 2, column 3, which is −4. The value 5 is a₁₃.

For the teacher: Understanding, low difficulty, standard 2.1.2

Textbook p. 38

Which pair of matrices is equal?

34 and 34
1234 and 1324
20 and 200
10.5 and 112

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Equal matrices need the same order and equal corresponding elements. Here both are 1 × 2 and 0.5 = 1/2. The others have different orders or different elements.

For the teacher: Understanding, medium difficulty, standard 2.1.3

Textbook p. 40

2−350 − −1426 =

3−73−6
1176
313−6
−37−36

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

For the teacher: Applying, medium difficulty, standard 2.2.1

Textbook p. 42

Calculate 32−104 + 10−25.

7−3−29
7−3−217
5−1−29
6−3012

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3A = 6−3012; adding gives 7−3−217. The last option forgot to add.

For the teacher: Applying, medium difficulty, standard 2.2.2

Textbook p. 46

Matrix A has order 3 × 2 and matrix B has order 2 × 4. What is the order of AB?

2 × 2
4 × 3
3 × 4
AB cannot be calculated

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Inner numbers match (2 = 2). The outer numbers give 3 × 4.

For the teacher: Understanding, medium difficulty, standard 2.2.3

Textbook p. 50

Which matrix is the identity matrix?

0110
1111
0000
1001

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The identity matrix has 1 on the main diagonal and 0 elsewhere. The third option is the zero matrix.

For the teacher: Remembering, low difficulty, standard 2.2.4

Textbook p. 55

123421 =

410
2234
410
37

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1(2) + 2(1) = 4 and 3(2) + 4(1) = 10.

For the teacher: Applying, medium difficulty, standard 2.2.3

Textbook p. 51

What is the determinant of 5231?

1
−1
−11
11

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ad − bc = 5(1) − 2(3) = 5 − 6 = −1.

For the teacher: Applying, medium difficulty, standard 2.2.5

Textbook p. 60

The matrix k346 has no inverse. Find k.

−2
8
2
0.5

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No inverse means ad − bc = 0: 6k − 12 = 0, so k = 2.

For the teacher: Analysing, high difficulty, standard 2.2.5

Textbook p. 60

Which matrix equation represents 2x − 3y = 5 and x + 4y = −1?

2314xy = 5−1
2−314xy = −15
21−34xy = 5−1
2−314xy = 5−1

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Coefficients of x form column 1 and of y column 2, in the order the equations are written, and the constants form the right side. The third option swaps rows and columns, and the first has the wrong sign for 3y.

For the teacher: Analysing, high difficulty, standard 2.2.6

Textbook p. 62

Section B: structured questions

34 marks

Answer all questions. Show your working.

The table shows the marks of Hana and Lim in a test.

	Malay	Mathematics	Science
Hana	78	85	90
Lim	66	92	74

[1 mark] (a)Write the marks as a matrix with one row for each pupil.

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788590669274 [1]1

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[2 marks] (b)State its order and say whether it is a square matrix.

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Order 2 × 3 [1]1
Not square: rows ≠ columns [1]1

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For the teacher: Applying, low difficulty, standard 2.1.1

Textbook p. 37

Given 2x − 153y8 = 9z + 2128, find the values of x, y and z.

[1 mark] (a)x

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2x − 1 = 9, x = 5 [1]1

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[1 mark] (b)y

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3y = 12, y = 4 [1]1

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[1 mark] (c)z

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z + 2 = 5, z = 3 [1]1

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For the teacher: Applying, medium difficulty, standard 2.1.3

Textbook p. 41

Given A = 4−216 and B = 30−52, calculate

[1 mark] (a)A + B

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7−2−48 [1]1

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[2 marks] (b)2A − B

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2A = 8−4212 [1]1
2A − B = 5−4710 [1]1

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For the teacher: Applying, medium difficulty, standard 2.2.2

Textbook p. 47

P = 3−124, Q = 2015 and R = 120−134.

[1 mark] (a)Can RP be calculated? Give a reason.

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No: R has 3 columns but P has only 2 rows [1]1

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[2 marks] (b)Calculate PQ.

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3(2) + (−1)(1)3(0) + (−1)(5)2(2) + 4(1)2(0) + 4(5) [1]1
5−5820 [1]1

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[2 marks] (c)Calculate PR.

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Order 2 × 3 [1]1
43−4−21616 [1]1

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For the teacher: Applying, medium difficulty, standard 2.2.3

Textbook p. 51

A canteen sells nasi lemak, roti canai and tea. The matrix S = 302540283235 gives the number sold on Monday (row 1) and Tuesday (row 2). The prices are RM2, RM1.50 and RM1 respectively, written as column matrix K.

[1 mark] (a)Write the matrix K and state the order of SK.

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K = 21.501; order 2 × 1 [1]1

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[2 marks] (b)Calculate SK.

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30(2) + 25(1.50) + 40(1)28(2) + 32(1.50) + 35(1) [1]1
137.50139 [1]1

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[1 mark] (c)On which day was the income higher, and by how much?

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Tuesday, by RM1.50 [1]1

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For the teacher: Analysing, medium difficulty, standard 2.2.3

Textbook p. 50

Given N = 5273.

[1 mark] (a)Calculate the determinant of N.

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5(3) − 2(7) = 1 [1]1

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[2 marks] (b)Find N⁻¹.

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N⁻¹ = (1/1)3−2−75 [1]1
= 3−2−75 [1]1

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[1 mark] (c)The matrix m436 has no inverse. Find m.

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6m − 12 = 0, m = 2 [1]1

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For the teacher: Applying, medium difficulty, standard 2.2.5

Textbook p. 60

2 pens and 3 notebooks cost RM12.60. 4 pens and 1 notebook cost RM9.20. Let a pen cost RMx and a notebook RMy.

[1 mark] (a)Write two linear equations for this information.

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2x + 3y = 12.60 and 4x + y = 9.20 [1]1

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[1 mark] (b)Write the equations as a matrix equation AX = B.

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2341xy = 12.609.20 [1]1

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[1 mark] (c)Find A⁻¹.

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|A| = 2(1) − 3(4) = −10, A⁻¹ = −1101−3−42 [1]1

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[2 marks] (d)Hence find the price of a pen and of a notebook.

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X = −1101−3−4212.609.20 = −110−15−32 [1]1
A pen costs RM1.50 and a notebook RM3.20 [1]1

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For the teacher: Applying, high difficulty, standard 2.2.7

Textbook p. 64

Aina found the inverse of C = 3152 and wrote C⁻¹ = 2153.

[1 mark] (a)Is her answer correct? Explain her mistake.

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No: she swapped 3 and 2 but did not change the signs of 1 and 5 [1]1

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[2 marks] (b)Write the correct C⁻¹.

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ad − bc = 3(2) − 1(5) = 1 [1]1
C⁻¹ = 2−1−53 [1]1

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[1 mark] (c)Show that CC⁻¹ = I.

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31522−1−53 = 1001 [1]1

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For the teacher: Evaluating, high difficulty, standard 2.2.5

Textbook p. 60

Make up two simultaneous linear equations in x and y whose solution is x = 2 and y = 3.

[2 marks] (a)Write the two equations.

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Any two independent equations satisfied by (2, 3), e.g. x + y = 5 [1]1
and x − y = −1 [1]1

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[1 mark] (b)Write them in the form AX = B and check that your |A| ≠ 0.

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111−1xy = 5−1; |A| = −2 ≠ 0 [1]1

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For the teacher: Creating, high difficulty, standard 2.2.6

Textbook p. 62

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Section B: structured questions

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20