5.1 Direct Variation (Part 1)

Posted on April 24, 2020 by user

(A) Determining whether a quantity varies directly as another quantity

1. If a quantity y varies directly as a quantity x, the

(a) y increases when x increases

(b) y decreases when x decreases

2. A quantity y varies directly as a quantity x if and only if

\frac{y}{x} = k

where k is called the constant of variation.

3. y varies directly as x is written as

y \propto x

.

4. When

y \propto x

, the graph of y against x is a straight line passing through the origin.

(B) Expressing a direct variation in the form of an equation involving two variables

Example 1

Given that y varies directly as x and y = 20 when x = 36 . Write the direct variation in the form of equation.

\begin{array}{l} y \propto x \\ y = k x \\ 20 = k (36) \\ k = \frac{20}{36} = \frac{4}{9} \to F i n d k f i r s t \\ ∴ y = \frac{4}{9} x \end{array}

5.3 Joint Variation

Posted on April 24, 2020 by user

5.3 Joint Variation

5.3a Representing a Joint Variation using the symbol ‘α’.

1. If one quantity is proportional to two or more other quantities, this relationship is known as joint variation.

2. ‘y varies directly as x and z’ is written as y α xz.

3. ‘y varies directly as x and inversely z’ is written as

y α \frac{x}{z} .

4. ‘y varies inversely as x and z’ is written as

y α \frac{1}{x z} .

Example 1:

State the relationship of each of the following variations using the symbol 'α'.

(a) x varies jointly as y and z.

(b) x varies inversely as y and

\sqrt{z} .

(c) x varies directly as r³ and inversely as y.

Solution:

\begin{array}{l} (a) x α y z \\ (b) x α \frac{1}{y \sqrt{z}} \\ (c) x α \frac{r^{3}}{y} \end{array}

5.3b Solving Problems involving Joint Variation

1. If

y α x^{n} z^{n}, then y = k x^{n} z^{n}

, where k is a constant and n = 2, 3 and ½.

2. If

y α \frac{1}{x^{n} z^{n}}, then y = \frac{1}{k x^{n} z^{n}}

, where k is a constant and n = 2, 3 and ½.

3. If

y α \frac{x^{n}}{z^{n}}, then y = \frac{k x^{n}}{z^{n}}

, where k is a constant and n = 2, 3 and ½.

Example 2:

Given that

p α \frac{1}{q^{2} \sqrt{r}}

when p = 4, q = 2 and r = 16, calculate the value of r when p = 9 and q = 4.

Solution:

\begin{array}{l} Given that p α \frac{1}{q^{2} \sqrt{r}}, p = \frac{k}{q^{2} \sqrt{r}} \\ When p = 4, q = 2 and r = 16, \\ 4 = \frac{k}{2^{2} \sqrt{16}} \\ 4 = \frac{k}{16} \\ k = 64 \\ ∴ p = \frac{64}{q^{2} \sqrt{r}} \\ When p = 9 and q = 4, \\ 9 = \frac{64}{4^{2} \sqrt{r}} \\ 9 = \frac{4}{\sqrt{r}} \\ \sqrt{r} = \frac{4}{9} \\ r = {(\frac{4}{9})}^{2} = \frac{16}{81} \end{array}

5.1 Direct Variation (Part 2)

Posted on April 24, 2020 by user

(C) Finding the value of a variable in a direct variation

1. When y varies directly as x and sufficient information is given, the value of y or x can be determined by using:

\begin{array}{l} (a) y = k x, or \\ (b) \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \end{array}

Example 2

Given that y varies directly as x and y = 24 when x = 8, find

(a) The equation relating y to x

(b) The value of y when x = 6

(c) The value of x when y = 36

Solution:

Method 1: Using y = kx

(a) y \propto x

y = kx

when y = 24, x = 8

24 = k (8)

k = 3

y = 3x

(b)
when x = 6,

y = 3 (6)

y = 18

(c)
when y = 36

36 = 3x

x =12

Method 2:

Using \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}}

(a)
Let x₁ = 8 and y₁= 24

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{y_{2}}{x_{2}} \\ \frac{3}{1} = \frac{y_{2}}{x_{2}} \to y_{2} = 3 x_{2} \\ ∴ y = 3 x \end{array}

(b)
Let x₁ = 8 and y₁= 24 and x₂= 6; find y₂.

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{y_{2}}{6} \\ y_{2} = \frac{24}{8} (6) \\ y_{2} = 18 \end{array}

(c)
Let x₁ = 8 and y₁= 24 and y₂= 36; find x₂.

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{36}{x_{2}} \\ 24 x_{2} = 36 \times 8 \\ x_{2} = 12 \end{array}

5.2 Inverse Variation

Posted on April 24, 2020 by user

5.2 Inverse Variation

If a quantity y varies inversely as another quantity x, then

(a) y increases when x decreases,

(b) y decreases when x increases

5.2b Expressing an inverse variation in the form of an equation

An inverse variation can be written in the form of an equation,

y = \frac{k}{x}

where k is a constant which can be determined.

Example 1:

Given y varies inversely as x and y = 4 when x =10. Write an equation which relates x and y.

Solution:

\begin{array}{l} y \propto \frac{1}{x} \to y = \frac{k}{x} \\ 4 = \frac{k}{10} \to k = 40 \\ ∴ y = \frac{40}{x} \end{array}

Example 2:

Given that y = 3 when x = 6, find the equation relates x and y if:

(a) y \propto \frac{1}{x^{2}} (b) y \propto \frac{1}{x^{3}} (c) y \propto \frac{1}{\sqrt{x}}

Solution:

\begin{array}{l} (a) y \propto \frac{1}{x^{2}} \to y = \frac{k}{x^{2}} \\ 3 = \frac{k}{6^{2}} \\ k = 3 \times 36 = 108 \\ ∴ y = \frac{108}{x^{2}} \\ (b) y \propto \frac{1}{x^{3}} \to y = \frac{k}{x^{3}} \\ 3 = \frac{k}{6^{3}} \\ k = 3 \times 216 = 648 \\ ∴ y = \frac{648}{x^{3}} \\ (c) y \propto \frac{1}{\sqrt{x}} \to y = \frac{k}{\sqrt{x}} \\ 3 = \frac{k}{\sqrt{6}} \\ k = 3 \times \sqrt{6} = 3 \sqrt{6} \\ ∴ y = \frac{3 \sqrt{6}}{\sqrt{x}} \end{array}

5.4 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

Question 1:

It is given that y varies directly as the cube of x and y = 192 when x = 4. Calculate the value of x when y = – 24.

Solution:

y α x³
y = kx³

192 = k (4)³

192 = 64 k

k = 3

y = 3 x³

when y = – 24

– 24 = 3 x³

x³ = – 8

x = – 2

Question 2:
It is given that y varies directly as the square of x and y = 9 when x = 2. Calculate the value of x when y = 16.

Solution:

y α x²
y = kx²
9 = k (2)²

\begin{array}{l} k = \frac{9}{4} \\ y = \frac{9}{4} x^{2} \\ When y = 16 \\ 16 = \frac{9}{4} x^{2} \\ x^{2} = \frac{64}{9} \\ x = \frac{8}{3} \end{array}

Question 3:
Given that y varies inversely as w and x and y = 45 when w = 2 and x =

\frac{1}{6}

. Find the value of x when y = 15 and w =

\frac{1}{3}

Solution:

\begin{array}{l} y α \frac{1}{w x} \\ y = \frac{k}{w x} \\ 45 = \frac{k}{(2) (\frac{1}{6})} \\ k = 45 \times \frac{1}{3} = 15 \\ y = \frac{15}{w x} \end{array}

\begin{array}{l} when y = 15, w = \frac{1}{3} \\ 15 = \frac{15}{(\frac{1}{3}) x} \\ \frac{x}{3} = 1 \\ x = 3 \end{array}

Question 4:
Given that

p α \frac{1}{q \sqrt{r}}

and p = 3 when q = 2 and r = 16, find the value of p when q = 3 and r = 4.

Solution:

\begin{array}{l} p α \frac{1}{q \sqrt{r}} \\ p = \frac{k}{q \sqrt{r}} \\ 3 = \frac{k}{(2) \sqrt{16}} \\ k = 24 \\ p = \frac{24}{q \sqrt{r}} \end{array}

\begin{array}{l} when q = 3, r = 4 \\ p = \frac{24}{3 \sqrt{4}} \\ p = \frac{24}{6} = 4 \end{array}

4.8 Solving Simultaneous Linear Equations using Matrices

Posted on April 24, 2020 by user

4.8 Solving Simultaneous Linear Equations using Matrices

1. Two simultaneous linear equations can be written in the matrix equation form.

For example, in the simultaneous equations:

ax + by = e

cx + dy = f

can be written in the matrix form as follows:

(\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} e \\ f \end{array}),

Where a, b, c, d, e and f are constant while x and y are unknowns.

Example 1:

Write the following simultaneous linear equations in the matrix form.

y– 6x – 19 = 0

2y + 3x + 22 = 0

Solution:

– 6x + y = 19

3x + 2y = – 22

The matrix form is:

(\begin{matrix} - 6 & 1 \\ 3 & 2 \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 19 \\ - 22 \end{array})

2. Matrix equations in the form

(\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} e \\ f \end{array})

can be solved for the unknowns x and y as follows.

(a) Let

A = (\begin{matrix} a & b \\ c & d \end{matrix})

and find A^-1.

(b) Multiply both sides of the equation by A^-1.

A^{- 1} (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = A^{- 1} (\begin{array}{l} e \\ f \end{array})

\begin{array}{l} (c) A^{- 1} A (\begin{array}{l} x \\ y \end{array}) = A^{- 1} (\begin{array}{l} e \\ f \end{array}) \\ I (\begin{array}{l} x \\ y \end{array}) = A^{- 1} (\begin{array}{l} e \\ f \end{array}) \\ ↑ \\ A^{- 1} A = I = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ (\begin{array}{l} x \\ y \end{array}) = A^{- 1} (\begin{array}{l} e \\ f \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{a d - b c} (\begin{matrix} d & - b \\ - c & a \end{matrix}) (\begin{array}{l} e \\ f \end{array}) \end{array}

Example 2:

Solve the following simultaneous equations by using the matrix method.

2x = 5 – 3y

7x = 1 – 5y

Solution:

2x + 3y = 5

7x + 5y = 1

\begin{array}{l} (\begin{matrix} 2 & 3 \\ 7 & 5 \end{matrix}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 5 \\ 1 \end{array}) \leftarrow \begin{array}{l} write the simultaneous \\ equations in matrix form . \end{array} \\ Let A = (\begin{matrix} 2 & 3 \\ 7 & 5 \end{matrix}) \\ A^{- 1} = \frac{1}{a d - b c} (\begin{matrix} d & - b \\ - c & a \end{matrix}) \\ A^{- 1} = \frac{1}{10 - 21} (\begin{matrix} 5 & - 3 \\ - 7 & 2 \end{matrix}) \\ A^{- 1} = \frac{1}{- 11} (\begin{matrix} 5 & - 3 \\ - 7 & 2 \end{matrix}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 11} (\begin{matrix} 5 & - 3 \\ - 7 & 2 \end{matrix}) (\begin{array}{l} 5 \\ 1 \end{array}) \leftarrow (\begin{array}{l} x \\ y \end{array}) = A^{- 1} (\begin{array}{l} e \\ f \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 11} (\begin{array}{l} 5 \times 5 + (- 3) \times 1 \\ - 7 \times 5 + 2 \times 1 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 11} (\begin{array}{l} 22 \\ - 33 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} - \frac{22}{11} \\ \frac{- 33}{- 11} \end{array}) = (\begin{array}{l} - 2 \\ 3 \end{array}) \\ ∴ x = - 2, y = 3. \end{array}

4.9 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

Question 1:

(\begin{matrix} 1 & 4 \\ 6 & 2 \end{matrix}) + 3 (\begin{matrix} 2 & 0 \\ 4 & - 3 \end{matrix}) - (\begin{matrix} - 3 & 0 \\ - 2 & - 5 \end{matrix})

Solution:

\begin{array}{l} (\begin{matrix} 1 & 4 \\ 6 & 2 \end{matrix}) + 3 (\begin{matrix} 2 & 0 \\ 4 & - 3 \end{matrix}) - (\begin{matrix} - 3 & 0 \\ - 2 & - 5 \end{matrix}) \\ = (\begin{matrix} 1 & 4 \\ 6 & 2 \end{matrix}) + (\begin{matrix} 6 & 0 \\ 12 & - 9 \end{matrix}) - (\begin{matrix} - 3 & 0 \\ - 2 & - 5 \end{matrix}) \\ = (\begin{matrix} 7 & 4 \\ 18 & - 7 \end{matrix}) - (\begin{matrix} - 3 & 0 \\ - 2 & - 5 \end{matrix}) \\ = (\begin{matrix} 7 - (- 3) & 4 - 0 \\ 18 - (- 2) & - 7 - (- 5) \end{matrix}) \\ = (\begin{matrix} 10 & 4 \\ 20 & - 2 \end{matrix}) \end{array}

Question 2:

Find the value of m in the following matrix equation:

(\begin{matrix} 9 & - 4 \\ 5 & 0 \end{matrix}) + \frac{1}{2} (\begin{matrix} 8 & m \\ 6 & - 10 \end{matrix}) = (\begin{matrix} 13 & - 7 \\ - 2 & 1 \end{matrix})

Solution:

\begin{array}{l} (\begin{matrix} 9 & - 4 \\ 5 & 0 \end{matrix}) + \frac{1}{2} (\begin{matrix} 8 & m \\ 6 & - 10 \end{matrix}) = (\begin{matrix} 13 & - 7 \\ - 2 & 1 \end{matrix}) \\ - 4 + \frac{1}{2} m = - 7 \\ \frac{1}{2} m = - 3 \\ m = - 6 \end{array}

Question 3:

\begin{array}{l} Given \\ (\begin{array}{l} 2 x \\ 3 y \end{array}) - 4 (\begin{array}{l} - 2 \\ 3 \end{array}) = (\begin{array}{l} - 2 \\ 6 \end{array}) \\ Find the values of x and y . \end{array}

Solution:

\begin{array}{l} 2 x + 8 = - 2 \\ 2 x = - 10 \\ x = - 5 \\ 3 y - 12 = 6 \\ 3 y = 18 \\ y = 6 \end{array}

Question 4:

\begin{array}{l} Given that matrix equation 3 (6 m) + n (3 4) = (12 7), \\ find the value of m + n \end{array}

Solution:

\begin{array}{l} 3 (6 m) + n (3 4) = (12 7) \\ 18 + 3 n = 12 \\ 3 n = - 6 \\ n = - 2 \\ 3 m + 4 n = 7 \\ 3 m + 4 (- 2) = 7 \\ 3 m = 15 \\ m = 5 \\ ∴ m + n = 5 + (- 2) = 3 \end{array}

4.10 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 1:

It is given that matrix A =

(\begin{matrix} 3 & - 1 \\ 5 & - 2 \end{matrix})

(a) Find the inverse matrix of A.

(b) Write the following simultaneous linear equations as matrix equation:

3u – v = 9

5u – 2v = 13

Hence, using matrix method, calculate the value of u and v.

Solution:
(a)

\begin{array}{l} A^{- 1} = \frac{1}{3 (- 2) - (5) (- 1)} (\begin{matrix} - 2 & 1 \\ - 5 & 3 \end{matrix}) \\ = - 1 (\begin{matrix} - 2 & 1 \\ - 5 & 3 \end{matrix}) = (\begin{matrix} 2 & - 1 \\ 5 & - 3 \end{matrix}) \end{array}

(b)

\begin{array}{l} (\begin{matrix} 3 & - 1 \\ 5 & - 2 \end{matrix}) (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} 9 \\ 13 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = - 1 (\begin{matrix} - 2 & 1 \\ - 5 & 3 \end{matrix}) (\begin{array}{l} 9 \\ 13 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = - 1 (\begin{array}{l} (- 2) (9) + (1) (13) \\ (- 5) (9) + (3) (13) \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = - 1 (\begin{array}{l} - 5 \\ - 6 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} 5 \\ 6 \end{array}) \\ ∴ u = 5, v = 6 \end{array}

Question 2:

It is given that matrix A =

(\begin{matrix} 2 & - 5 \\ 1 & 3 \end{matrix})

and matrix B =

m (\begin{matrix} 3 & k \\ - 1 & 2 \end{matrix})

such that AB =

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

(a) Find the value of m and of k.

(b) Write the following simultaneous linear equations as matrix equation:

2u – 5v = –15

u + 3v = –2

Hence, using matrix method, calculate the value of u and v.

Solution:

(a) Since AB =

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

, B is the inverse of A.

m = \frac{1}{(2) (3) - (- 5) (1)} = \frac{1}{11}

k = 5

(b)

\begin{array}{l} (\begin{matrix} 2 & - 5 \\ 1 & 3 \end{matrix}) (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} - 15 \\ - 2 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = \frac{1}{11} (\begin{matrix} 3 & 5 \\ - 1 & 2 \end{matrix}) (\begin{array}{l} - 15 \\ - 2 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = \frac{1}{11} (\begin{array}{l} (3) (- 15) + (5) (- 2) \\ (- 1) (- 15) + (2) (- 2) \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = \frac{1}{11} (\begin{array}{l} - 55 \\ 11 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} - 5 \\ 1 \end{array}) \\ ∴ u = - 5, v = 1 \end{array}

4.4 Multiplication of Matrix by a Number

Posted on April 24, 2020 by user

4.4 Multiplication of a Matrix by a Number

When matrix is multiplied by a number, every element in the matrix is multiplied by the number.

Example:

Given that

A = (\begin{matrix} - 2 & 4 \\ 5 & - 6 \end{matrix})

, find each of the following.
(a) 3A
(b) -2A

Solution:

\begin{array}{l} (a) 3 A = 3 (\begin{matrix} - 2 & 4 \\ 5 & - 6 \end{matrix}) \\ = (\begin{matrix} 3 \times (- 2) & 3 \times 4 \\ 3 \times 5 & 3 \times (- 6) \end{matrix}) \\ = (\begin{matrix} - 6 & 12 \\ 15 & - 18 \end{matrix}) \end{array}

\begin{array}{l} (b) - 2 A = - 2 (\begin{matrix} - 2 & 4 \\ 5 & - 6 \end{matrix}) \\ = (\begin{matrix} - 2 \times (- 2) & - 2 \times 4 \\ - 2 \times 5 & - 2 \times (- 6) \end{matrix}) \\ = (\begin{matrix} 4 & - 8 \\ - 10 & 12 \end{matrix}) \end{array}

4.4 Multiplication of Matrix by a Number (Sample Questions)

Posted on April 24, 2020 by user

Example 1:

Express the following as single matrix.

(a) 3 (\begin{matrix} 1 & - 4 \\ - 3 & 2 \end{matrix}) + 2 (\begin{matrix} - 3 & 1 \\ 3 & - 4 \end{matrix})

(b) 2 (\begin{matrix} - 1 & 0 \\ 9 & - 4 \end{matrix}) - 4 (\begin{matrix} - 2 & 2 \\ 3 & - 6 \end{matrix}) - \frac{1}{3} (\begin{matrix} - 9 & - 3 \\ - 6 & 15 \end{matrix})

Solution:

\begin{array}{l} (a) 3 (\begin{matrix} 1 & - 4 \\ - 3 & 2 \end{matrix}) + 2 (\begin{matrix} - 3 & 1 \\ 3 & - 4 \end{matrix}) \\ = (\begin{matrix} 3 \times 1 & 3 \times (- 4) \\ 3 \times (- 3) & 3 \times 2 \end{matrix}) + (\begin{matrix} 2 \times (- 3) & 2 \times 1 \\ 2 \times 3 & 2 \times (- 4) \end{matrix}) \\ = (\begin{matrix} 3 & - 12 \\ - 9 & 6 \end{matrix}) + (\begin{matrix} - 6 & 2 \\ 6 & - 8 \end{matrix}) \\ = (\begin{matrix} 3 + (- 6) & - 12 + 2 \\ - 9 + 6 & 6 - (- 8) \end{matrix}) \\ = (\begin{matrix} - 3 & - 10 \\ - 3 & 14 \end{matrix}) \end{array}

\begin{array}{l} (b) \\ 2 (\begin{matrix} - 1 & 0 \\ 9 & - 4 \end{matrix}) - 4 (\begin{matrix} - 2 & 2 \\ 3 & - 6 \end{matrix}) - \frac{1}{3} (\begin{matrix} - 9 & - 3 \\ - 6 & 15 \end{matrix}) \\ = (\begin{matrix} - 2 & 0 \\ 18 & - 8 \end{matrix}) - (\begin{matrix} - 8 & 8 \\ 12 & - 24 \end{matrix}) - (\begin{matrix} \frac{1}{3} \times (- 9) & \frac{1}{3} \times (- 3) \\ \frac{1}{3} \times (- 6) & \frac{1}{3} \times 15 \end{matrix}) \\ = (\begin{matrix} - 2 & 0 \\ 18 & - 8 \end{matrix}) - (\begin{matrix} - 8 & 8 \\ 12 & - 24 \end{matrix}) - (\begin{matrix} - 3 & - 1 \\ - 2 & 5 \end{matrix}) \\ = (\begin{matrix} - 2 - (- 8) - (- 3) & 0 - 8 - (- 1) \\ 18 - 12 - (- 2) & - 8 - (- 24) - 5 \end{matrix}) \\ = (\begin{matrix} 9 & - 7 \\ 8 & 11 \end{matrix}) \end{array}