4.2 Equal Matrices (Sample Questions)

Posted on April 24, 2020 by user

Example 1:

State the values of the unknowns in the following pairs of equal matrix.

(\begin{matrix} 1 & x + 2 \\ 4 - y & - 1 \end{matrix}) = (\begin{matrix} 1 & 3 \\ 2 & - 1 \end{matrix})

Solution:

(\begin{matrix} 1 & x + 2 \\ 4 - y & - 1 \end{matrix}) = (\begin{matrix} 1 & 3 \\ 2 & - 1 \end{matrix})

x + 2 = 3
x = 1

4 – y = 2
–y = –2
y = 2

Example 2:

Calculate the values of p and q in each of the following matrix equations.

\begin{array}{l} (a) (\begin{matrix} 3 & 2 p + q \\ p & - 3 \end{matrix}) = (\begin{matrix} 3 & 1 \\ 8 - 2 q & - 3 \end{matrix}) \\ (b) (\begin{matrix} 10 & 0 \\ 5 p - 8 & 1 \end{matrix}) = (\begin{matrix} p - 2 q & 0 \\ - 4 q & 1 \end{matrix}) \end{array}

Solution:

(a) (\begin{matrix} 3 & 2 p + q \\ p & - 3 \end{matrix}) = (\begin{matrix} 3 & 1 \\ 8 - 2 q & - 3 \end{matrix})

2p + q = 1

q = 1 – 2p ----(1)

p = 8 – 2q ----(2)

Substitute (1) into (2),

p = 8 – 2(1 – 2p)

p = 8 – 2 + 4p

p – 4p = 6

–3p = 6

p = –2

Substitute p = –2 into (1),

q = 1 – 2(–2)

q = 5

(b) (\begin{matrix} 10 & 0 \\ 5 p - 8 & 1 \end{matrix}) = (\begin{matrix} p - 2 q & 0 \\ - 4 q & 1 \end{matrix})

10 = p – 2q

p = 10 + 2q ----(1)

5p – 8 = –4q ----(2)

Substitute (1) into (2),

5 (10 + 2q) – 8 = –4q

50 + 10q – 8 = –4q

14q = –42

q = –3

Substitute q = –3 into (1),

p = 10 + 2(–3)

p = 4

4.3 Addition and Subtraction of Matrices

Posted on April 24, 2020 by user

4.3 Addition and Subtraction of Matrices

(A) Determining whether addition or subtraction can be performed on two given matrices

1. Two matrices can be added or subtracted if both matrices have the same order.

2. The addition or subtraction of two matrices is performed by adding or subtracting the corresponding elements of the matrices.

Example 1:

State whether the following matrices can be added or subtracted. Give reason to your answer.

\begin{array}{l} (a) (\begin{array}{l} 2 \\ 3 \end{array}) and (1 8) \\ (b) (\begin{matrix} 1 & 2 \\ 7 & 1 \end{matrix}) and (\begin{matrix} 10 & 0 \\ - 3 & - 1 \end{matrix}) \end{array}

(c) (p 2 4) and (2 6 q)

Solution:
(a) Cannot because the matrices are not of the same order.
(b) Can because the matrices are of the same order.
(c) Can because the matrices are of the same order.

Example 2:

Express each of the following matrices as a single matrix.

\begin{array}{l} If A = (\begin{array}{l} 3 \\ 2 \end{array}) and B = (\begin{array}{l} 5 \\ 1 \end{array}), then \\ (a) A + B = (\begin{array}{l} 3 \\ 2 \end{array}) + (\begin{array}{l} 5 \\ 1 \end{array}) \\ = (\begin{array}{l} 3 + 5 \\ 2 + 1 \end{array}) \\ = (\begin{array}{l} 8 \\ 3 \end{array}) \end{array}

\begin{array}{l} (b) B - A = (\begin{array}{l} 3 \\ 2 \end{array}) - (\begin{array}{l} 5 \\ 1 \end{array}) \\ = (\begin{array}{l} 3 - 5 \\ 2 - 1 \end{array}) \\ = (\begin{array}{l} - 2 \\ 1 \end{array}) \end{array}

SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Further Practice:

Transformation III, Long Questions (Question 1)

Question 1:

(a) Transformation T is a translation

(\begin{array}{l} - 4 \\ 2 \end{array})

and transformation P is an anticlockwise rotation of 90^oabout the centre (1, 0).

State the coordinates of the image of point (5, 1) under each of the following transformation:

(i) Translation T,

(ii) Rotation P,

(iii) Combined transformation T².

(b) Diagram below shows three quadrilaterals, ABCD, EFGH and JKLM, drawn on a Cartesian plane.

(i) JKLM is the image of ABCD under the combined transformation VW.

Describe in full the transformation:

(a) W (b) V

(ii) It is given that quadrilateral ABCD represents a region of area 18 m².

Calculate the area, in m², of the region represented by the shaded region.

Solution:

(a)

(b)

(i)(a)

W: A reflection in the line x = –2

(i)(b)

V: An enlargement of scale factor 3 with centre (0, 4).

(b)(ii)

Area of EFGH = area of ABCD = 18 m²

Area of JKLM = (Scale factor)² × Area of object

= 3² × area of EFGH

= 3² × 18

= 162 m²

Therefore,

Area of the shaded region

= Area of JKLM – area of EFGH

= 162 – 18

= 144 m²

SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 2:
(a) Diagram below shows point A and straight line y + x = 5 drawn on a Cartesian plane.

Transformation T is a translation

(\begin{array}{l} 5 \\ - 2 \end{array})

Transformation R is a reflection at the line y + x = 5.

State the coordinates of the image of point A under each of the following transformations:
(i) Transformation T,
(ii) Combined transformation TR.

(b) Diagram below shows pentagons JKLMN, PQRST and PUVWX, drawn on a Cartesian plane.

(i) PUVWX is the image of JKLMN under the combined transformation CB.
Describe in full the transformation:
(a) B (b) C

(ii) It is given that pentagon JKLMN represents a region of area 80 m² .
Calculate the area, in m² , of the region represented by the shaded region.

Solution:
(a)

(i) (3, 4) → T → (8, 2)
(ii) (3, 4) → R → (1, 2) → T → (6, 0)

(b)

(b)(i)(a)
B: A clockwise rotation of 90^o about the centre (0, 2).

(b)(i)(b)

\begin{array}{l} Scale factor = \frac{P U}{P Q} = \frac{6}{4} = \frac{3}{2} \\ C: An enlargement of scale factor \frac{3}{2} with centre P (2, 0) \end{array}

(b)(ii)
Area of PQRST = Area of JKLMN
= 80 m²

\begin{array}{l} Area of P U V W X \\ = {(\frac{3}{2})}^{2} \times area of P Q R S T \\ = \frac{9}{4} \times 80 \\ = 180 m^{2} \\ ∴ Area of the shaded region \\ = area of P U V W X - area of P Q R S T \\ = 180 - 80 \\ = 100 m^{2} \end{array}

3.1b Determining the Image of an Object under Combination of (a) Two Enlargements or (b) an Enlargement and an Isometric Transformation

Posted on April 24, 2020 by user

3.1b Determining the Image of an Object under Combination of (a) Two Enlargements or (b) an Enlargement and an Isometric Transformation

1. Enlargement is a transformation where all points of an object on a plane move from a fixed point at a constant ratio.

2. The fixed point is called the centre of enlargement and constant ratio is called the scale factor.

Scale factor, k = \frac{Length of image side}{Length of object side}

3. For enlargement, the object and the image are similar.

4. Area of image = (Scale factor)² × Area of object

= k² × Area of object

Example:

E, P and T are three transformations that are defined as follows:

E = Enlargement with centre V (0, –1) and a scale factor of 2.

P = Reflection at the line y = –1.

T = T r a n s l a t i o n (\begin{array}{l} 3 \\ - 3 \end{array})

Based on the diagram above, determine the image of the shaded figure under the combined transformations

(a) E² (b) ET (c) EP

Solution:

(a)

Shaded figure → (E) figure III → (E) figure D.

Hence, the image of the shaded figure under the combined transformation E²= EE is the figure D.

(b)

Shaded figure → (T) figure II → (E) figure A.

Hence, the image of the shaded figure under the combined transformation ET is the figure A.

(c)

Shaded figure → (P) figure I → (E) figure B.

Hence, the image of the shaded figure under the combined transformation EP is the figure B.

3.1d Describing the Combination of Two Transformations Given the Object and the Image

Posted on April 24, 2020 by user

3.1d Describing the Combination of Two Transformations Given the Object and the Image

Example 1:

In the diagram above, triangle KLM is the image of triangle ABC under the combined transformations VW. Describe in full, the transformation W and the transformation V.

Solution:

Figure ABC → W→ Figure KB’C’ → V → Figure KLM

W = Translation (\begin{array}{l} 0 \\ 4 \end{array})

V= Enlargement with centre K(1, 1) and a scale factor of 3.

Example 2:

In the diagram above, triangle CKL is the image of triangle ABC under the combined transformations UY. Describe in full, the transformation Y and the transformation U.

Solution:

Figure ABC → Y→ Figure AB’C → U → Figure CKL

Y = Reflection in the straight line y = 6

U = Enlargement with centre C(12, 6) and a scale factor of 2.

Example 3:

In the diagram above, triangle PQR is the image of triangle ABC under the combined transformations TW. Describe in full, the transformation W and the transformation T.

Solution:

Figure ABC → W→ Figure PBC’ → T → Figure PQR

W = Clockwise rotation of 90^o about the point B (5, 5)

T = Enlargement with centre P (1, 5) and a scale factor of 2.

3.2 Solving Problems involving Transformations

Posted on April 24, 2020 by user

3.2 Solving Problems involving Transformations

Example:

The diagram below shows the triangles ABC, SQT and PQR.

(a) Transformation T is a translation

(\begin{array}{l} 5 \\ - 2 \end{array})

Transformation P is a reflection in the straight line y = 7.

State the coordinates of the image of point A under each of the following transformations:

(i) Transformation T,

(ii) Combined transformation PT.

(b)

(i) PQR is the image of triangle ABC under the combined transformations MN. Describe in full, the transformation N and the transformation M.

(ii) Given that the area of the shaded region PSTR is 200 m², calculate the area of the triangle SQT.

Solution:
(a)

(i) A(3, 11) → T → A’(8, 9).

(ii) A(3, 11) → T → A’(8, 9) → P → A”(8, 5).

(b)(i)

N = Anticlockwise rotation of 90^o about the point B(5, 10).

(b)(ii)

M = Enlargement with centre (5, 12) and a scale factor of 3.

(c)

Area of image = (Scale factor)² x Area of object

Area of PQR = 3² x Area of SQT

Area of SQT + Area of PSTR = 9 x Area of SQT

Area of SQT + 200 = 9 x Area of SQT

200 = 9 x Area of SQT – Area of SQT

200 = 8 x Area of SQT

Area of S Q T = \frac{200}{8}

Area of SQT = 25 m²

3.1 Combination of Two Transformations

Posted on April 24, 2020 by user

3.1a Determining the image of an object under the combination of two isometric transformations

1. Translations, reflections and rotations are isometric transformations.

2. In an isometric transformation, the shape and the size of the image is the same as the object.

3. For two transformations, A and B,

(a) Combined transformation AB represents transformation B followed by transformation A.

(b) Combined transformation BA represents transformation A followed by transformation B.

(c) Combined transformation A²= AA represents transformation A which is carried out twice consecutively.

3.1c Stating the Coordinates of the Image of a Point under a Combined Transformation

Posted on April 24, 2020 by user

3.1c Stating the Coordinates of the Image of a Point under a Combined Transformation

1. The coordinates of the image of a point, K, under the combined transformation AB can be determined by the following steps.

Step 1:

Determine the coordinates of K’, the image of K, under the first transformation, B.

Step 2:

Determine the coordinates of K”, the image of K’, under the second transformation, A. K” is the image of K, under the combined transformation, AB.

Example:

T, P, R and E are four transformations that are defined as follows:

T = T r a n s l a t i o n (\begin{array}{l} - 4 \\ 3 \end{array})

P = Reflection in the y–axis.

R = Clockwise rotation 90^o about the origin

E = Enlargement with scale factor of 3 and the origin as centre.

Find the coordinates of the image of the point A (3, –2) under each of the following combined transformation.

(a) TT (b) PT (c) ET (d) ER (e) EP

Solution:

(a)

A(3, –2) → T → A’(–1, 1) → T → A”(–5, 4).

(b)

A(3, –2) → T → A’(–1, 1) → P → A”(1, 1).

(c)

A(3, –2) → T → A’(–1, 1) → E → A”( –3, 3).

(d)

A(3, –2) → R → A’(–2, 3) → E → A”( –6, –9).

(e)

A(3, –2) → P → A’(–3, –2) → E → A”( –9, –6).

2.5 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 6:
(a) Complete the table in the answer space for the equation

y = \frac{36}{x}

by writing down the values of y when x = 3 and x = 8.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 1 cm to 1 unit on the x-axis and 1 cm to 1 unit on the y-axis, draw the graph of

y = \frac{36}{x}

for 2 ≤ x ≤ 14.

(c) From your graph, find
(i) the value of y when x = 2.6,
(ii) the value of x when y = 4.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation

\frac{36}{x} + x - 14 = 0

for 2 ≤ x ≤ 14.

Answer:

x	2	2.4	3	4	6	8	10	12	14
y	18	15		9	6		3.6	3	2.6

Solution:
(a)

\begin{array}{l} y = \frac{36}{x} \\ when x = 3 \\ y = \frac{36}{3} = 12 \\ when x = 8 \\ y = \frac{36}{8} = 4.5 \end{array}

(b)

(c)
(i) From the graph, when x = 2.6, y = 13.6
(ii) From the graph, when y = 4, x = 9

(d)

\begin{array}{l} y = \frac{36}{x} ........... (1) \\ 0 = \frac{36}{x} + x - 14 ........... (2) \\ (1) - (2) : \\ y = - x + 14 \end{array}

The suitable straight line is y = –x + 14.

x	2	12
y = –x + 14	12	2

From the graph, x = 3.4, 10.6.