Application of Algebra

Logical reasoning is the foundation of sound mathematical thinking. It helps us form clear arguments and test whether conclusions follow from given statements. This section explores simple and compound statements and how we use truth tables to determine the validity of arguments. As we study logic, we will notice its strong link to Set theory studied in Year 1; especially in how union, intersection, and complement relate to logical connectors like “and”, “or”, and “not”. In fact, De Morgan’s Laws, covered in Year 2, which describe how set operations behave under complements, also apply directly to logic. Understanding these connections strengthens our reasoning skills and prepares us to handle more complex mathematical ideas with confidence and precision.

KEY IDEAS

• Compound statements – statements which are formed by combining two or more simple statements using logical connectives. • Mathematical statements – statements which have a definite state of being either True or False. They cannot be both True and False. The statements may contain symbols or words. • Truth table – a chart which uses logic to show all possible truth values of a statement or combination of statements. It helps us determine whether a compound statement is always true, sometimes true, or never true and whether an argument is valid. • Validity of an argument – In logic, an argument is valid if, and only if, all the premises are true, then the conclusion must also be true.

In our communication with others, we make statements all the time. Sometimes they are true, other times they are not. We make use of some information to determine whether the statements are true or not. In mathematics, we have “Mathematical Statements”. These are sentences that have a definite state of being, either True or False. They cannot be both True and False. The statement may contain symbols or words. This implies that expressions whose truth value cannot be determined are not statements. If we cannot tell whether it is True or False, then it is not a statement. Let us look at the sentence “3 is an odd number”. Based on our knowledge of odd numbers, we can conclude that the sentence is true. Hence, it is a mathematical statement.

Example 1.1

Determine whether the following are True or False.

1. $\pi$ is a rational number

2. If the product of two numbers is 0, then at least one of the numbers must be 0.

3. All parallelograms are rectangles

4. A number is divisible by 3 if the sum of its digits is divisible by 3.

5. 27 is a prime number

6. $\pi + 3 \geq 6$

7. The sum of the interior angles of a triangle is $240^\circ$ .

Solution

1. False
2. True
3. False
4. True
5. False
6. True
7. False

Notation of Statements

Mathematical statements are denoted by capital letters such as A, B, D, K, etc. D: $18 - 5$ = 13, simply means D is the statement $18 - 5$ = 13. D is obviously a true statement because $18 - 5$ = 13.

Example 1.2

Express the following as notations of statements using any variable of your choice and determine their truth value.

1. $\sqrt{7}$ is an irrational number.
2. $11 + \sqrt{4} \in E$ where, $E$ is the set of even numbers.
3. $23 \in Q$ where, $Q$ is the set of irrational numbers.
4. All integers are real numbers.

Solution

1. $\mathrm{A}: \sqrt{7}$ is an irrational number. (True)
2. $\mathrm{G}: 11 + \sqrt{4} \in \mathrm{E}$ where, $\mathrm{E}$ is the set of even numbers. (False)
3. $\mathrm{K}: 23 \in \mathrm{Q}$ where, $\mathrm{Q}$ is the set of irrational numbers. (False)
4. $\mathrm{Y}:$ All integers are real numbers. (True)

Negation of a Statement

Assuming that $Q$ is a statement, the negation of $Q$ , denoted by $\sim Q$ , is also a statement claiming the opposite truth value to $Q$ . This means that if $Q$ is true then $\sim Q$ is false. On the other hand, if $Q$ is false then $\sim Q$ is true. Let us go through this activity on negation of statements.

Activity 1.1 Exploring negation of statements

1. Refer to the statements below a. $\mathrm{A}: \sqrt{7}$ is an irrational number. (True) b. $\mathrm{G}: 11 + \sqrt{4} \in \mathrm{E}$ where, $\mathrm{E}$ is the set of even numbers. (False) c. $\mathrm{K}: 23 \in \mathrm{Q}$ where, $\mathrm{Q}$ is the set of irrational numbers. (False) d. $\mathrm{Y}:$ All integers are real numbers. (True)
2. To negate statements $1\mathrm{a} - 1\mathrm{d}$ , change all positive statements to the negative and vice versa. a. $\sim \mathrm{A}: \sqrt{7}$ is not an irrational number. b. $\sim \mathrm{G}: 11 + \sqrt{4} \notin \mathrm{E}$ where, $\mathrm{E}$ is the set of even numbers. c. $\sim \mathrm{K}: 23 \notin \mathrm{Q}$ where, $\mathrm{Q}$ is the set of irrational numbers. d. $\sim \mathrm{Y}:$ All integers are not real numbers.
3. Determine the truth value of the negated statements. a. $\sim \mathrm{A}: \sqrt{7}$ is not an irrational number. (False) b. $\sim \mathrm{G}: 11 + \sqrt{4} \notin \mathrm{E}$ where, $\mathrm{E}$ is the set of even numbers. (True) c. $\sim \mathrm{K}: 23 \notin \mathrm{Q}$ where, $\mathrm{Q}$ is the set of irrational numbers. (True) d. $\sim \mathrm{Y}:$ All integers are not real numbers. (False)
4. Now, negate the already negated statements in Step 3 (double negation). a. $\sim(\sim \mathrm{A}): \sqrt{7}$ is an irrational number. b. $\sim(\sim \mathrm{G}): 11 + \sqrt{4} \in \mathrm{E}$ where, $\mathrm{E}$ is the set of even numbers. c. $\sim(\sim \mathrm{K}): 23 \in \mathrm{Q}$ where, $\mathrm{Q}$ is the set of irrational numbers. d. $\sim(\sim \mathrm{Y}):$ All integers are real numbers.
5. Discuss your observations on the truth value of the original statement and the negated statement as well as the outcome of the double negation.

From the observation, you will realise that negating the original statement leads to a change in the truth value, where False becomes True and vice versa. Also, the double negation statements were the same as the original.

Double Negation

From Activity 1.1 , the truth value of $\mathrm{G}$ is the opposite of the truth value of $\sim \mathrm{G}$ and the effect of taking double negation, that is $\sim$ ( $\sim \mathrm{G}$ ), was the same as that of $\mathrm{G}$ . Here we would say the statements $\sim$ ( $\sim \mathrm{G})$ and $\mathrm{G}$ are the same. However, in some cases, where statements $Q$ and $\sim$ ( $\sim Q$ ) are not identical as statements, but have the same truth values for all possible choices of statements $Q$ , we say that $Q$ and $\sim$ ( $\sim Q$ ) are logically equivalent, or that $Q$ is logically equivalent to $\sim$ ( $\sim Q$ )and use the notation $\sim(\sim Q) \equiv Q$ . In everyday use, an example can be: $Q:$ All triangles have three sides $\sim Q:$ All triangles do not have three sides $\sim$ ( $\sim Q$ ): It is not true that all triangles do not have three sides. Here $Q$ and $\sim$ ( $\sim Q$ ) are not worded identically but hold the same truth value hence $\sim(\sim Q) \equiv Q$ .

Statements and Venn diagrams

So far, we have expressed statements using sentences and symbols. It is important to note that statements can be expressed as sets and subsequently represented on Venn diagrams. Recall set representation on Venn diagrams learnt in Year 1.

Example 1.3

Given the statement “M: Ignatius is a Computing student”, identify the Universal set, Subsets and Complements that can be created from the statement and represent it on a Venn diagram.

Solution

Let $U$ represent the Universal Set Let $C$ represent the Set of computing students Let $G$ represent Ignatius $U =$ {All students} $C =$ {Computing students} $G =$ Ignatius Now, $G \in C$

$FIG$

Example 1.4

Given the statement “T: 18 is a multiple of 3, but 7 is not”, identify the Universal set, Subsets and Complements that can be created from the statement and represent it on a Venn diagram.

Solution

Let $\mathrm{U}$ represent the Universal Set Let $\mathrm{R}$ represent the Set of multiples of 3 Let $\mathrm{R}'$ represent non-multiples of 3 $U =$ {Real numbers} Now, $18 \in \mathrm{R}$ and $7 \in \mathrm{R}'$ . $FIG$

Compound Statements

We have been talking about statements for a while now, some of the statements we have come across have connectives (and, or, then, but) while others have no connectives. The statements made by combining two or more connectives are known as compound statements. Those without connectives are called simple statements. Different connecting words yield different compound statements. For instance, the word AND is used for Conjunction, OR is used for Disjunction, IF-THEN for implication, etc. Let us go through a few examples of Compound Statements.

Example 1.5

1. 39 is an odd number and 8 is an even number
2. If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
3. A set of real numbers is made up of rational and irrational numbers.
4. If a number is divisible by 5 then it ends with a 0 or a 5. Note A compound statement can have more than 1 logical connector.

In logical reasoning, drawing implications involves identifying what conclusions follow from certain statements. This includes working with implicative statements, exploring their converses, and applying the chain rule of implications to connect multiple ideas. Mastering these concepts helps in forming clear, valid arguments and understanding how truth flows through connected reasoning.

Implicative Statement

In English Language, we learn about conditional sentences and the various types. Conditional sentence Type 1 is the “If” statement. An example is “If Kwame goes to school, he will excel in his academics”. In mathematics, that condition helps us generate what we call an Implicative Statement. It is a type of compound statement that establishes a conditional relationship between two simple statements using the “if-then” structure and is a fundamental concept in mathematical logic. Remember, the statements can be true or false. Let us go through some examples.

Example 1.6

1. If it is a prime number then, it is an odd number.
2. If three angles are supplementary then, their sum equals $180^\circ$ .
3. If $\sqrt[3]{v} = 2$ then $v = 8$ .
4. If a polygon is three-sided then, it is a triangle. There is a symbolic way of representing implicative statements. That is two statements $M$ and $N$ expressed as “If $M$ then $N$ ”, can be written symbolically as $M \Longrightarrow N$ . $M$ and $N$ are referred to as the hypothesis and conclusion respectively. Let us go through some examples to express statements symbolically

Example 1.7

Express the following statements symbolically.

1. If a polygon is a square, then it is a rectangle.
2. If a number is a multiple of 10, then it is also a multiple of 5.
3. If a matrix is invertible, then its determinant is not zero.
4. If a number is prime, then it has exactly two factors.

Solution

1. a polygon is a square $\Longrightarrow$ a polygon is a rectangle
2. Let $M_{10}(x): x$ a multiple of 10 $M_5(x): x$ a multiple of 5 Then symbolically, $M_{10}(x) \Longrightarrow M_5(x)$ .
3. Let $Inv(B):$ matrix $B$ is invertible $\det(\mathrm{B}):$ determinant of matrix $B$ Then symbolically, $Inv(B) \Longrightarrow (\det(B) \neq 0)$ .
4. Let $P(x)$ : $x$ is prime $F(x):$ number of factors of $x$ Then symbolically, $P(x) \Longrightarrow (F(x) = 2)$

Converse of an implication

Recall in implicative statements we have the “hypothesis” and “conclusion”? To create the converse of an implication, we switch the initial positioning of the “hypothesis” and “conclusion”. Let us look at the implicative statements in Example 1.6 , and identify the hypothesis and conclusion of each statement. Table 1.1: Identifying hypothesis and conclusion Hypothesis Conclusion It is a prime number It is an odd number. Three angles are supplementary Their sum equals $180^\circ$ . $\sqrt[3]{v} = 2$ $v = 8$ . A polygon is three-sided It is a triangle. To create converses of these statements we shall make the conclusion become the hypothesis and vice versa as seen in Table 1.2.

Table 1.2: Identifying hypothesis and conclusion Hypothesis Conclusion It is an odd number. It is a prime number Their sum equals $180^\circ$ . Three angles are supplementary $v = 8$ . $\sqrt[3]{v} = 2$ It is a triangle. A polygon is three-sided Therefore, the converses of the initial implications will read:

1. If it is an odd number, then it is a prime number.
2. If the sum of three angles equals $180^\circ,$ then they are supplementary.
3. If $v = 8$ , then $\sqrt[3]{v} = 2.$
4. If it is a triangle, then it is a polygon which is three-sided. Note You have to adjust the tenses and sentence structure where necessary as done for the second and fourth statements . To show the repositioning of the hypothesis and conclusion symbolically, the converse may be represented using “ $\Longleftarrow$ ”.
5. It is a prime number $\Longleftarrow$ it is an odd number.
6. Three angles are supplementary $\Longleftarrow$ their sum equals $180^\circ$ .
7. $\sqrt[3]{v} = 2$ $\Longleftarrow$ $v = 8$ .
8. A polygon is three-sided $\Longleftarrow$ it is a triangle.

The chain rule of implication

A logical principle that provides a connection of multiple implicative statements to form a new valid statement is what we call “The chain rule of implication”. Given the two implicative statements Q and T: Q: If a number is divisible by 9, then it is divisible by 3. T: If a number is divisible by 3, then it is divisible by 1. The chain-rule of implication will be; If a number is divisible by 9 then it is divisible by 1. Let us go through more examples. 1 q qaA Q1

Example 1.8

Express the following as a chain of implications

1. If a figure is a square then it is a rectangle. If it is a rectangle then it has 4 sides.
2. If $y = 4$ then $y^2 = 16$ . If $y^2 = 16$ then $y^2 - 16 = 0$ .

Solution

1. If a figure is a square then it has four sides. $FIG$
2. If $y = 4$ then $y^2 - 16 = 0$ .

In mathematics, we often work with ideas that can be stated as true or false. Before exploring how these ideas can be combined, it is important to first understand what simple statements are and how they relate to compound statements. This foundation will prepare us to see how logical connectives bring statements together to form meaningful expressions.

Identifying Simple Statements

A simple statement is a basic declarative statement with no logical connectives such as AND, OR, IF – THEN, NOT and IF AND ONLY IF. Such statements cannot be broken down into smaller logical sentences and express only a single, complete thought. Examples of simple statements include:

1. The sun is a star.
2. 17 is a prime number
3. London is the capital city of Britain
4. Tamale is the capital town of Northern Region
5. $\sqrt{11}$ is an irrational number.

Forming Compound Statements Using Connectives

Earlier we mentioned connectives (AND, OR, IF-THEN, etc.) and their relevance in compound statements. To form a compound statement, we combine two simple statements using any of the logical connectives. Let us explore using the statements R and V R: 17 is a prime number V: $\sqrt{9}$ is a rational number. An example of a compound statement will be “17 is a prime number and $\sqrt{9}$ is a rational number”.

Activity 1.2 Forming Compound Statements

Step 1 : Provide 5 simple mathematical statements Step 2 : Using the statements created in Step 1, form 5 compound statements using any logical connective of your choice.

We have talked about the compound statements: Conjunction, Disjunction and Implication (Conditional). We will now examine their symbolic representations while also looking at Bi-conditional compound statements. • Conjunction statements (AND): When two simple statements are combined with the logical connective “AND” we create a compound statement known as a conjunction statement. Given two simple statements $E$ and $R$ combined using the 'AND' connective, the compound statement can be written symbolically as $E \wedge R$ . Note For a conjunction compound statement, both simple statements must be true for the compound statement to be true.

Example 1.9

1. $\sqrt{169}$ is 13 and $\sqrt[3]{64}$ is 4. Symbolically written as $\sqrt{169} = 13$ $\wedge$ $\sqrt[3]{64} = 4$ .
2. $y$ is greater than 2 and less than 10 simultaneously Symbolically can be expressed as $(y > 2) \wedge (y < 10)$
3. A number is even and is a multiple of 4. • Disjunction statements (OR) : When two simple statements are combined with the logical connective “OR” we create a compound statement known as a Disjunction

statement. The two simple statements represented as $E$ and $R$ can be combined using the OR connective and are written as $E \vee R$ . Note In a disjunction statement, any of the simple statements must be true for the disjunction compound statement to be true.

Example 1.10

1. $n$ is either less than 0 or greater than 15. Symbolically $(n < 0) \vee (n > 15)$
2. $g$ is 6 or $g$ is 18 Symbolically, $(g = 6) \vee (g = 18)$
3. A number is prime or it is even. • Conditional Statement (IF-THEN) : The connective used for a conditional statement is if - then. If Judith does well in the test, then she will be promoted to the next class. Here, the first statement, $E$ , can be taken as the hypothesis, and the second statement, $R$ , can be taken as the conclusion. We can express conditional statements of these two simple statements $E$ , $R$ as “If $E$ then $R$ ”. Note The conditional compound statement does not hold true if the hypothesis is true and the conclusion is false. But in all other situations, the conditional statement is true.

Example 1.11

1. If $a$ and $b$ are odd integers, then $a + b$ is even. Symbolically, $[(a \bmod 2 = 1) \wedge (b \bmod 2 = 1)] \Longrightarrow [(a + b) \bmod 2 = 0$ ]
2. If $n$ is a prime number greater than 2, then $n$ is odd. Symbolic form: $[n > 2 \wedge \mathrm{Prime}(n)] \Rightarrow [n \bmod 2 = 1]$
3. If the gradient of a line is $0$ , then the line is horizontal. Let $m$ be the gradient of the line Symbolically, ( $m = 0) \Longrightarrow$ Horizontal line • Bi-conditional Statement (IF AND ONLY IF) : The bi-conditional statement uses the connective 'If and only if,' which is represented by the symbol $\Leftrightarrow$ .The two statements, $E$ and $R$ , are represented as a compound statement, $E$ $\Leftrightarrow$ $R$ , and here, the first statement, $E$ , is called the hypothesis (antecedent), and the second statement, $R$ , is called the conclusion (consequent). Note The bi-conditional compound statement is true if both statements are either true or both

are false.

Example 1.12

1. A triangle is equilateral if and only if all its interior angles are $60^\circ$ Symbolically, $[$ Equilateral $(\triangle)] \Leftrightarrow$ $[$ All angles $= 60^\circ$ ]
2. An integer is even if and only if it is divisible by 2. Symbolically, $(n$ is even) $\Leftrightarrow (n \bmod 2 = 0$ )
3. Two figures are congruent if and only if they have the same size and shape. Let the figures be labelled A and B Symbolically, Congruent (A, B) $\Leftrightarrow$ (Same size $\wedge$ Same shape)

Truth Table of a Compound Statement

In simple statements, it is fairly easy to determine the truth value. We only need to critically analyse the statement and check if a counter example exists to determine whether the original statement is true or false. However, for a compound statement, you need to analyse the truth value of the individual simple statement then factor in the logical connective being used to determine the truth value. A more convenient way to establish the truth value of a compound statement is by making use of a truth table. We shall discuss the truth table representations for each on the logical connectives mentioned earlier.

Conjunction Truth Table

To create a conjunction truth table, we consider the situations where both simple statements are true, one of the statements is false and when both statements are false. Based on these, four different pairings (orderings) can be generated as shown in Table 1.4 for two simple statements $D$ and $K$ . Table 1.4: Pairings for two statements $\mathrm{D}$ , $\mathrm{K}$ . $\mathbf{D}$ $\mathbf{K}$ T T T F F T F F Table 1.5: Truth table for $\mathrm{D} \wedge \mathrm{K}$ $\mathbf{D}$ $\mathbf{K}$ $\mathbf{D} \wedge \mathbf{K}$ T T T T F F F T F F F F Observations from Table 1.5

1. For row 1 in Table 1.5, both statements $(\mathrm{D}$ , $\mathrm{K})$ are true. Since “AND” requires both to be true, the result is true.
2. For row 2, statement D is true but statement K is false. However, “AND” needs both statements to be true resulting in a false result.
3. For row 3, statement D is false but statement K is true. However, “AND” needs both statements to be true resulting in a false result.
4. For row 4 in Table 1.5, both statements $(\mathrm{D}$ , $\mathrm{K})$ are false. Since “AND” requires both to be true, the result is false. In summary, the reasons for the compound statements being false is because one or both of the statements is false. We can have a conjunction table for three or more statements. For instance, given three simple statements $D, K, M$ , we will have; Sample Space for $(D, K, M)$ $\mathbf{D}$ $\mathbf{K}$ $\mathbf{M}$ T T T T T F T F T T F F F T T F T F F F T F F F The conjunction truth table for the three statements $\mathbf{D}$ $\mathbf{K}$ $\mathbf{M}$ $\mathbf{D} \wedge \mathbf{K}$ $\mathbf{D} \wedge \mathbf{K} \wedge \mathbf{M}$ T T T T T T T F T F T F T F F T F F F F F T T F F F T F F F F F T F F F F F F F Remember that for a conjunction compound statement to be True, both statements must be true .

Disjunction Truth Table

This uses the connective 'or' to form a compound statement. Here, even if only one of the individual statements is true, the compound statement holds true. Table 1.6: Truth table for $\mathbf{D} \vee \mathbf{K}$ $\mathbf{D}$ $\mathbf{K}$ $\mathbf{D} \vee \mathbf{K}$ Reasoning Both statements are true. “OR” only needs at least one true statement T T T thus the result is true. Since at least one of the statements is true, the resulting statement is T F T true Since at least one of the statements is true, the resulting statement is F T T true F F F Since both statements are false, the resulting statement is false. Exploring a Disjunction table for three statements, we will have: Sample Space for $(D, K, M)$ $\mathbf{D}$ $\mathbf{K}$ $\mathbf{M}$ T T T T T F T F T T F F F T T F T F F F T F F F The disjunction truth table for the three statements $\mathbf{D}$ $\mathbf{K}$ $\mathbf{M}$ $\mathbf{D} \vee \mathbf{K}$ $\mathbf{D} \vee \mathbf{K} \vee \mathbf{M}$ T T T T T T T F T T T F T T T T F F T T F T T T T F T F T T F F T F T F F F F F Remember that for a disjunction compound statement to be True , one or both individual statements must be true .

Implication truth table

This uses the If-then connective, which is represented as $\Rightarrow$ . Here, the statement $\mathbf{D}$ is referred to as the hypothesis, the statement $\mathbf{K}$ is referred to as the conclusion. The compound statement is true if the conclusion is true, irrespective of the hypothesis. Also, the compound statement is true if both the hypothesis and the conclusion are false. Table 1.7 shows the representation of the implicative connective, $\mathbf{D} \Rightarrow \mathbf{K}$ . Table 1.7: Truth table for $\mathbf{D} \Rightarrow \mathbf{K}$ $\mathbf{D}$ $\mathbf{K}$ $\mathbf{D} \Rightarrow \mathbf{K}$ Reasoning If the hypothesis is true and the conclusion is true, then the T T T implication is satisfied. This is the only case where the implication is false because we T F F claimed $\mathrm{K}$ must be true if $\mathrm{D}$ is true, but $\mathrm{K}$ turned out false. If the hypothesis is false, the implication is considered true F T T regardless of the conclusion; we didn’t make any real promise to break. Again, since the hypothesis is false, the implication doesn't make F F T any false promise. So, the statement holds true.

Extended Activity

Create an implication truth table for three statements by yourself to show how well you have understood the concept.

Bi-conditional (If and only if) truth table

This uses the connective 'if and only if' and is represented as $\Leftrightarrow$ . Given the statements $\mathbf{D}$ and $\mathbf{K}$ , the first statement, $\mathbf{D}$ , is referred to as antecedent, and the second statement, $\mathbf{K}$ , is referred to as consequent. The validity of each of statement can be either true or false. The bi-conditional compound statement is true if both first and second statements have the same truth values and is false otherwise. Displaying the if and only if connective, $\mathbf{D} \Leftrightarrow \mathbf{K}$ in a table. Table 1.8: Truth table for $\mathbf{D} \Leftrightarrow \mathbf{K}$ $\mathbf{D}$ $\mathbf{K}$ $\mathbf{D} \Leftrightarrow \mathbf{K}$ T T T T F F F T F F F T Let us go through some examples showing how we build truth tables using connectives

Example 1.13

If $p$ and $q$ are two simple statements such that $p$ is true and $q$ is false, construct the truth table for a. $\sim p \wedge q$ b. $\sim(p \vee q)$

Solution

Truth table for $\sim p \wedge \sim q$ $p$ $q$ $\sim p$ $\sim p \wedge q$ T F F F Truth table for $\sim(p \vee \sim q)$ $p$ $q$ $p \vee q$ $\sim(p \vee q)$ T F T F

Valid and invalid arguments using the Truth table

Using truth tables, the validity of an argument can be checked by inspecting the premises and conclusion in the rows. If all the premises in the row are true and the conclusion is also true, then the argument will be valid. On the other hand, if all the premises are true and the conclusion is false, then the argument is invalid. Note A row in a truth table where all the premises are true is called a critical row .

Example 1.14

Refer to the statements below: M: If a number is divisible by 4, then it is an even number. N: 16 is divisible by 4 Q: 16 is an even number. a. Represent the argument using symbols: $\mathrm{A}:$ A number is divisible by $4$ $\mathrm{B}:$ The number is even b. Use a truth table to show whether this argument is valid or not.

Solution

a. “If a number is divisible by 4, then it is an even number” can be written as: If $A$ then $B$ , $\mathrm{A} \Rightarrow \mathrm{B}$ . From the original statements:

1. $\mathrm{A} \Rightarrow \mathrm{B}$
2. $\mathrm{A}$ is true
3. So, $\mathrm{B}$ is true. b. To test the validity of the argument let us create a truth table Table 1.19: Truth Table for $\boldsymbol{A} \Rightarrow \boldsymbol{B}$ $\mathbf{A}$ $\mathbf{B}$ $\mathbf{A} \Rightarrow \mathbf{B}$ T T T T F F F T T F F T Let us rearrange the entries so the premises come before the conclusion Table 1.20: Rearranged Truth Table for $\boldsymbol{A} \Rightarrow \boldsymbol{B}$ $\mathbf{A}$ $\mathbf{A} \Rightarrow \mathbf{B}$ $\mathbf{B}$ T T T T F F F T T F T F The premises are: “a number is divisible by 4”, “If a number is divisible by 4, then it is an even number”. The conclusion is “16 is an even number”. From Table 1.20, both premises are true and the conclusion is also true. Hence the argument is valid.

Example 1.15

Consider the following statements: J: If it is raining, then Ama will carry an umbrella K: Ama did not carry an umbrella L: Therefore, it is not raining. a. Represent the argument using symbols if: G : It is raining H: Ama will carry an umbrella b. Use a truth table to test whether this argument is valid or not.

Solution

a. Now, $G \Rightarrow H \equiv J$ Then Premise 1 is $\mathrm{G} \Rightarrow \mathrm{H} \equiv \mathrm{J}$ (If it is raining, then Ama will carry an umbrella) Premise 2 $\sim \mathrm{H} \equiv \mathrm{K}$ (Ama did not carry an umbrella) Conclusion $\sim \mathrm{G} \equiv \mathrm{L}$ (It is not raining) The argument is $\mathrm{G} \Rightarrow \mathrm{H}$ , $\sim \mathrm{H}$ , $\sim \mathrm{G}$ . b. Using the truth table, we have: $\mathbf{G}$ $\mathbf{H}$ $\mathbf{G} \Rightarrow \mathbf{H}$ $\sim \mathbf{H}$ $\sim \mathbf{G}$ T T T F F T F F T F F T T F T F F T F T From the truth table, $\mathrm{G} \Rightarrow \mathrm{H}$ is True, $\sim \mathrm{H}$ is False and $\sim \mathrm{G}$ is False, in row 2. In row 4, $\mathrm{G} \Rightarrow \mathrm{H}$ is True, $\sim \mathrm{H}$ is False and $\sim \mathrm{G}$ is True. Hence, we can conclude that the argument is not valid. Recall that for an argument to be valid, all the premises in the row must be true as well as the conclusion.

Activity 1.3 Validity of Arguments

1. Choose a motion on the topic “Mathematics should be taught as a language rather than a science”.
2. Prepare a half-page write-up in support of the motion you choose. Include examples and references to support your claim.
3. Take turns presenting your team’s argument.
4. Identify the simple, compound statements and the logical connectives in your write-up.
5. Now, the opposing team should respond by stating the negation of the presented statements and evaluate whether the argument is logically valid or not.
6. After all statements have been debated, both teams should work together to construct at least three valid chain rule implications using any of the statements debated. Example

a. If mathematics uses symbols, structure and rules of grammar, then mathematics functions like a language. b. If mathematics functions like a language, then it should be taught using methods suited for teaching language. c. If mathematics should be taught using methods suited for teaching language, then mathematics should be taught as a language rather than as a science. 7. Present your final chain rule implications to your classmates. Seek help from your teacher to confirm if the final chain rule implications are logically valid using truth tables or standard reasoning.

1. A school policy states: If a student fails to attend class regularly, then the student will not be allowed to sit the final exam. If a student is not allowed to sit the final exam, then the student cannot pass the course. a. Construct the symbolic form of this argument and apply the chain rule of implications to derive a new implication. b. Using a truth table, determine whether the new implication is valid and always true. Justify your conclusion.
2. Consider the two compound statements: a. Construct a truth table for both and . b. Determine whether and are logically equivalent. Justify your conclusions using observations from the truth table. c. Explain how the response in (b) relates to De Morgan’s Laws.
3. A learner makes the following argument: If I study hard, I will pass the test. I passed the test. Therefore, I must have studied hard. a. Express this argument in symbolic form. b. Construct a truth table for this argument. c. Determine whether the argument is valid or invalid using truth table analysis. d. Explain the flaw in the learner’s reasoning, if any.
4. Let $\mathrm{P}$ : “The student submitted the assignment.” Let $\mathrm{Q}$ : “The student passed the course.” Let $\mathrm{R}$ : “The student graduated.” $\mathrm{P} \Rightarrow \mathrm{Q}$ $\mathrm{Q} \Rightarrow \mathrm{R}$ You are told that: $\sim \mathrm{R}$ : The student did not graduate a. Using the chain rule of implication, represent a single implication involving $\mathrm{P}$ and $\mathrm{R}$ . b. Construct a truth table for the resulting implication in (a) and $\sim \mathrm{R}$ . c. Based on your truth table, determine if it is logically possible that the student submitted the assignment. Justify your answer using reasoning from the table.

In this section on matrices and linear transformations, we will explore how matrices can serve as essential tools for solving linear equations and representing transformations. Matrices enable us to tackle real-world problems in various fields, such as engineering, economics, and computer graphics. By understanding matrix inverses, we can efficiently encode and decode messages, highlighting their significance in cryptography. We aim to equip you with the skills to apply matrices in practical scenarios and solve related problems. Ultimately, understanding these concepts will enhance your mathematical abilities and prepare you for advanced applications in mathematics.

KEY IDEAS

• Composition of Transformations: The composition of two transformations P and Q is denoted as P ∘ Q and is computed as the product of their matrices: PQ. • Finding the Image : Given an object point and a transformation matrix, the image can be calculated by matrix multiplication. • Inverse Transformations: The inverse of a transformation reverses the mapping, returning points to their original positions. • Linear transformation : A linear transformation maps points (objects) in a plane to new points (images) using a matrix. • Transformations include : Translation, rotation, reflection, and enlargement.

Concept of matrices in transformation

One critical area for the application of matrices is linear transformation, which represents the one-to-one mapping of a point called the object in a plane into a point called the image in the plane. The object is the original point, while the image is the point obtained. For example, if point $A(1, 2)$ is transformed to $A'(-2, 1),$ The object is taken as $A$ and $A'$ is the image. An object may undergo translation, rotation, reflection and enlargement. Details of these transformations: translation, rotation, reflection and enlargement will be discussed in the later part of this section.

Activity 2.1 Writing linear transformation in matrix form

1. Consider the linear transformation $x' = px + qy$ and $y' = mx + ny$ .
2. Write a matrix for the linear transformation by writing the coefficient of $x$ and $y$ .
3. Discuss your result with your peers or teacher. Note • Given the linear transformation, $x' = px + qy$ and $y' = mx + ny$ , we write the matrix as $\begin{pmatrix} p & q \\ m & n \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ • $x'$ and $y'$ are the transformed coordinates • $x$ and $y$ are the original coordinates • $p, q, m$ , and $n$ are constants that define the transformation.

Example 2.1

Find the matrix of the linear transformation $x' = 4x + 2y$ and $y' = 3x - 5y$

Solution

Write down the coefficient of x and y in both equations $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 3 & -5 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ Hence, the transformation matrix is $\begin{pmatrix} 4 & 2 \\ 3 & -5 \end{pmatrix}$

Example 2.2

Determine the matrix of the linear transformation $x' = x - y$ and $y' = 10x + 3y$

Solution

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 10 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ Hence, the transformation matrix is $\begin{pmatrix} 1 & -1 \\ 10 & 3 \end{pmatrix}.$

Example 2.3

Determine the matrix of the linear transformation $x' = y$ and $y' = 2x$

Solution

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ Hence, the transformation matrix is $\begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix}$

Example 2.4

Determine the matrix for the linear transformation $P: (x, y) \rightarrow \left(\frac{1}{2}x + 6y, 12x + 2\right)$ Solution $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 6 \\ 12 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ Hence, the transformation matrix is $\begin{pmatrix} \frac{1}{2} & 6 \\ 12 & 2 \end{pmatrix}$

Equation of the image of a line under a given

transformation

To find the equation of the image of a line under a translation, reflection, rotation, or enlargement, go through the following steps:

1. Find the coordinates of any two points on the line by choosing 2 numbers for x, and substituting into the equation to find the corresponding y values.
2. Find the coordinates of the images of the two points from Step 1.
3. Use the two image points from Step 2 to find the slope of the image line.
4. Substitute one of the image points and the slope from step 3 into the equation of a line formula to determine the equation of the image line.

Example 2.5

Find the equation of the image of the line $y = 3x + 4$ , under the transformation $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ .

Solution

Step 1: Find any two points that lie on line $y = 3x + 4$ If $x = 0$ , $y = 3(0) + 4$ $y = 4$

$A(0, 4)$ If $x = 2$ $y = 3(2) + 4$ $y = 10$ $B(2, 10)$ Hence, two points that lie on line $y = 3x + 4$ are $A(0, 4)$ and $B(2, 10)$ . Step 2: Add the points to the translation vector to obtain the image $A' = \begin{bmatrix} 5 \\ 2 \end{bmatrix} + \begin{bmatrix} 0 \\ 4 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$ $B' = \begin{bmatrix} 5 \\ 2 \end{bmatrix} + \begin{bmatrix} 2 \\ 10 \end{bmatrix} = \begin{bmatrix} 7 \\ 12 \end{bmatrix}$ The image of these points under the translation $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ will be $\mathrm{A}'$ (5, 6) and $B'$ (7, 12). Step 3: Find gradient using the image points; $m_1 = \frac{12 - 6}{7 - 5}$ $m_1 = 3$ Step 4 : Substitute $m_1$ into $y - y_1 = m_1(x - x_1)$ . $y - 6 = 3(x - 5)$ $y - 6 = 3x - 15$ $y = 3x - 9$ Thus, the equation of the image line is $y = 3x - 9$ .

Example 2.6

The line $y = 2x + 5$ is transformed by a translation vector $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ find the image line.

Solution

Find any two points on the line using two arbitrary values of $x$ . $x = 0$ , $y = 2(0) + 5 \qquad (0, 5)$ $x = 2$ , $y = 2(2) + 5 \qquad (2, 9)$ Find the image points by adding the translation vector $\begin{pmatrix} 0 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}$ $\begin{pmatrix} 2 \\ 9 \end{pmatrix} + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 12 \end{pmatrix}$ Find the gradient using the image points $\frac{12 - 2}{4 - 8} = -2.5$ Use one of the image points to form the equation of the image line $y - 8 = -2.5(x - 2)$ $y = -2.5x + 13$

Example 2.7

The line $y = 4x + 1$ has been transformed by the linear equations $x' = 2x - 3y$ and $y' = 5x + 2y$ . Find the equation of the resulting line.

Solution

Step 1 : Write the equations as a matrix. $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ The matrix is $\begin{pmatrix} 2 & -3 \\ 5 & 2 \end{pmatrix}$ Step 2 : Find two points that lie on the line $y = 4x + 1$ . If $x = 0,$ $y = 4(0) + 1$ $y = 1$ $A(0, 1)$ If $x = -3$ $y = 4(-3) + 1$ $y = -11$ $B(-3, -11)$ The two points are $\mathrm{A}(0, 1)$ and $\mathrm{B}(-3, -11)$ Step 3 : Determine the images of the points through matrix multiplication $A' = \begin{pmatrix} 2 & -3 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ $= \begin{pmatrix} 2(0) - 3(1) \\ 5(0) + 2(1) \end{pmatrix}$ $= \begin{pmatrix} -3 \\ 2 \end{pmatrix}$ $B' = \begin{pmatrix} 2 & -3 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} -3 \\ -11 \end{pmatrix}$ $= \begin{pmatrix} 2(-3) - 3(-11) \\ 5(-3) + 2(-11) \end{pmatrix}$ $= \begin{pmatrix} 27 \\ -37 \end{pmatrix}$ The image of these points under the $\begin{pmatrix} 2 & -3 \\ 5 & 2 \end{pmatrix}$ will be $\mathrm{A}'(-3, 2)$ and $B'(27, -37).$ Step 4 : Find gradient using the image points; $m_1 = \frac{-37 - 2}{27 - (-3)}$ $m_1 = -\frac{13}{10}$

Step 5 : Substitute $m_1$ into $y - y_1 = m_1(x - x_1)$ . $y - 2 = -\frac{13}{10}(x - (-3))$ $y - 2 = -\frac{13}{10}(x + 3)$ $10y - 20 = -13x - 39$ The equation of the resulting line is $10y + 13x + 19 = 0$ .

We can use a linear transformation to find the image when we know the object. To do this, assume we want to find the image of Q (x, y) under the matrix $A =$ $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , go through the activity below.

Activity 2.2 How to find the image when given the object

1. Identify the given point, object or position vector, $\begin{bmatrix} x \\ y \end{bmatrix}$ .
2. Identify the mapping represented by the 2 × 2 matrix, $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ .
3. Multiply the position vector by the mapping to obtain the image of the point: $\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ = $\begin{bmatrix} ax + by \\ cx + dy \end{bmatrix}$

Example 2.8

Find the image of Q (-2, 3) under the matrix A= $\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

Solution

The object is given by Q (-2, 3) Multiply the given matrix by the vector, $OQ' = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} -2 \\ 3 \end{bmatrix}$ $= \begin{bmatrix} 1(-2) + 3(3) \\ 2(-2) + 4(3) \end{bmatrix}$ $= \begin{bmatrix} -2 + 9 \\ -4 + 12 \end{bmatrix} = \begin{bmatrix} 7 \\ 8 \end{bmatrix}$ Therefore, $Q'(7, 8)$ , is the image of Q (-2, 3) under the matrix A. Note Write the image in row form, just as the object was written.

Example 2.9

Find the image of the point P (4, -6) under the transformation associated with the matrix $\begin{bmatrix} 3 & 10 \\ -6 & 14 \end{bmatrix}$ .

Solution

$\begin{bmatrix} 3 & 10 \\ -6 & 14 \end{bmatrix}\begin{bmatrix} 4 \\ -6 \end{bmatrix}$ = $\begin{bmatrix} 3(4) + 10(-6) \\ -6(4) + 14(-6) \end{bmatrix}$ = $\begin{bmatrix} 12 - 60 \\ -24 - 84 \end{bmatrix}$ = $\begin{bmatrix} -48 \\ -108 \end{bmatrix}$ Thus, the image of P is $P'(-48, -108)$

Example 2.10

A point $S(a, b)$ was transformed by the matrix $\begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}$ to the image $S'(5, 9)$ Find S.

Solution

Using the same procedure as in the other examples, $\begin{bmatrix} 5 \\ 9 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix}$ $\begin{bmatrix} 5 \\ 9 \end{bmatrix} = \begin{bmatrix} 1(a) - 2(b) \\ 3(a) + 0(b) \end{bmatrix}$ $\begin{bmatrix} 5 \\ 9 \end{bmatrix} = \begin{bmatrix} a - 2b \\ 3a \end{bmatrix}$ $5 = a - 2b \ldots\ldots\ldots (1)$ $9 = 3a$ ………..(2) $\frac{9}{3} = \frac{3a}{3}$ $a = 3$ $5 = 3 - 2b$ $5 - 3 = -2b$ $\frac{2}{-2} = \frac{-2b}{-2}$ $b = -1$ Hence $S(3, -1)$

Transformation of a given matrix

In pairs, or individually, write down the transformation matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ in linear transformation form. The answer is $T(x, y) \rightarrow (ax + by, cx + dy)$ .

To transform the matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ to a linear transformation, we multiply the matrix by the vector $\begin{bmatrix} x \\ y \end{bmatrix}$ . i.e., $\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} ax + by \\ cx + by \end{bmatrix}$ Writing this in row form gives $T(x, y) \rightarrow (ax + by, cx + dy)$ .

Example 2.11

Write down the following matrices in linear transformation form a. $\begin{bmatrix} 2 & -4 \\ 9 & -1 \end{bmatrix}$ b. $\begin{bmatrix} \frac{1}{4} & 3 \\ 0 & 5 \end{bmatrix}$ c. $\begin{bmatrix} 2\sqrt{2} & \sqrt{3} \\ \sqrt{3} & 5\sqrt{2} \end{bmatrix}$

Solution

a. $\begin{bmatrix} 2 & -4 \\ 9 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ $\begin{bmatrix} 2x - 4y \\ 9x - y \end{bmatrix}$ $T(x, y) \rightarrow (2x - 4y, 9x - y)$ b. $\begin{bmatrix} \frac{1}{4} & 3 \\ 0 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ $\begin{bmatrix} \frac{1}{4}x + 3y \\ 5y \end{bmatrix}$ $T(x, y) \rightarrow \left(\frac{1}{4}x + 3y, 5y\right)$ c. $\begin{bmatrix} 2\sqrt{2} & \sqrt{3} \\ \sqrt{3} & 5\sqrt{2} \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ $\begin{bmatrix} 2\sqrt{2}x + \sqrt{3}y \\ \sqrt{3}x + 5\sqrt{2}y \end{bmatrix}$ $T(x, y) \rightarrow (2\sqrt{2}x + \sqrt{3}y, \quad \sqrt{3}x + 5\sqrt{2}y)$

Geometric Transformation

A geometric transformation is a way to change a shape or figure in space. This change can affect the position, size, shape , or orientation of the figure. There are four main types of geometric transformations:

1. Translation : Moving a shape from one place to another without changing its size or shape.
2. Reflection : Flipping a shape over a line, like looking in a mirror.
3. Rotation : Turning a shape around a point, like spinning a wheel.
4. Dilation : Changing the size of a shape while keeping its proportions the same, mak- ing it larger or smaller.

Example 2.12

Given that A = $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ , describe the linear transformation $T_A$ geometrically

Solution

Consider the starting points of (1, 0) and (0, 1). The x coordinate has been unchanged as it remains (1, 0) as indicated by the first column. The second column indicates that the point (0, 1) is now at (0, -1) which shows that it has been reflected in the $x$ axis. This means that the linear transformation $T_A$ represented by the matrix A geometrically performs a reflection in the $x -$ axis.

Example 2.13

Describe the linear transformation T M geometrically of the matrix M = $\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$ .

Solution

This matrix represents a dilation (scaling) with a scale factor of 5. Geometric Interpretation: Every point moves away from the origin, making the shape larger while preserving its proportions.

Example 2.14

Describe the linear transformation $T_M$ geometrically of the matrix M = $\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$

Solution

This matrix represents a rotation by $180^\circ$ about the origin. It negates both the x and y coordinates. Geometric Interpretation: Every point moves to the opposite quadrant.

Example 2.15

Describe the linear transformation $T_M$ geometrically of the matrix $M =$ $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

Solution

Transformation: This matrix represents a reflection in the line $y = x$ . It swaps the x and y coordinates. Geometric Interpretation: Points are reflected in the line $y = x$ .

Composition of linear transformation

Imagine you have two different linear transformations, represented by two different matrices, P and Q. If you first apply the transformation represented by matrix Q, and then immediately apply the transformation represented by matrix P, this combined transformation is called P ∘ Q (read as P composed with Q). Mathematically, we can write this as $\mathrm{P} \circ \mathrm{Q} = \mathrm{PQ}$ , which means that the combined trans- formation is equal to the product of the two matrices, P and Q. Similarly, $\mathrm{Q} \circ \mathrm{P}$ (P followed by Q) is given by the product of $QP$ , i.e., Q ∘ P = QP To calculate the composite transformation A ∘ B:

1. Write the matrices A and B that represent the individual linear transformations: a. Matrix A represents the first transformation. b. Matrix B represents the second transformation.
2. Find the matrix product AB: a. Multiply the matrices B and A in that order. b. The order is important because the second transformation (B) is applied first, followed by the first transformation (A).
3. Write the linear transformation for AB: a. The matrix AB represents the combined transformation, where the second transformation (B) is applied first, followed by the first transformation (A).

b. The combined transformation can be written as $\mathrm{AB}$ ( $x, y),$ where ( $x, y$ ) are the input coordinates. 4. Observe the order of the matrices: a. When carrying out the matrix product, the matrix that represents the first transformation (A) is placed on the left, and the matrix that represents the second transformation (B) is placed on the right. b. This order reflects the fact that the second transformation (B) is applied first, followed by the first transformation (A).

Example 2.16

If M = $\begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix}$ and N = $\begin{bmatrix} 11 & 6 \\ 3 & -1 \end{bmatrix}$ , evaluate: i. $\mathrm{M} \circ \mathrm{N}$ ii. $\mathrm{N} \circ \mathrm{M}$ iii. Comment on your results in (i) and (ii).

Solution

i. M ∘ N = MN $= \begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix}\begin{bmatrix} 11 & 6 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} 2(11) + 3(3) & 2(6) + 3(-1) \\ 5(11) + 4(3) & 5(6) + 4(-1) \end{bmatrix}$ $= \begin{bmatrix} 22 + 9 & 12 - 3 \\ 55 + 12 & 30 - 4 \end{bmatrix}$ $= \begin{bmatrix} 31 & 9 \\ 67 & 26 \end{bmatrix}$ ii. N ∘ M = $\begin{bmatrix} 11 & 6 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix}$ $= \begin{bmatrix} 11(2) + 6(5) & 11(3) + 6(4) \\ 3(2) - 1(5) & 3(3) - 1(4) \end{bmatrix}$ $= \begin{bmatrix} 22 + 30 & 33 + 24 \\ 6 - 5 & 9 - 4 \end{bmatrix}$ $= \begin{bmatrix} 52 & 57 \\ 1 & 5 \end{bmatrix}$ iii. Since M ∘ N and N ∘ M are not equal, they are not commutative, confirming that Matrix multiplication is not commutative, thus $M \circ N \neq N \circ M$ .

Activity 2.3 Finding the Translation of a Linear Transformation

1. Identify the points $A$ and $B$ plotted on the graph in Figure 2.1.
2. What are the coordinates of points $A$ and $B$ ?
3. Starting at point A, perform the following activity Given the vector $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ , i. Count 1 step to the right of A Remember that positive numbers on the x-axis are counted to the right, while negative numbers are counted to the left ii. Count 2 steps down from where you reached in 3(ai.) Positive numbers on the $y -$ axis are counted upwards while negative numbers are counted downwards
4. Plot a point at the new destination and name it $A'$ (…, …)
5. Repeat the steps from 3 to 4 for point $B$
6. What did you obtain for point $B'$ ? [Figure] Figure 2.1: Translation of a vector Points A and B are the objects, while $A'$ and $B'$ are the images obtained when they were translated by the vector $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ . The other way to obtain the images is given by Image of object $=$ Object + Translation vector Let us use this to find the images of A and B to confirm your results in the activity above

A is point (4, 4) and is translated by the vector $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ , Image $= \begin{pmatrix} 4 \\ 4 \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$ Thus, $A'(5, 2)$ B is point (-6, 5) and is translated by the vector $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ Image $= \begin{pmatrix} -6 \\ 5 \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ $= \begin{pmatrix} -5 \\ 3 \end{pmatrix}$ Hence, $\mathrm{B}'(-5, 3)$

Example 2.17

A triangle ABC has the vertices $A(3, 4), B(-1, 5)$ and $C(0, 0)$ . Triangle $ABC$ is translated by the vector $\begin{pmatrix} -2 \\ -3 \end{pmatrix}$ , find the coordinate of the image triangle $\mathrm{A'B'C'}$ .

Solution

Take each object and add the translated vector to it For A, the image $A' = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \begin{pmatrix} -2 \\ -3 \end{pmatrix}$ $= \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ For B, the image $B' = \begin{pmatrix} -1 \\ 5 \end{pmatrix} + \begin{pmatrix} -2 \\ -3 \end{pmatrix}$ $= \begin{pmatrix} -3 \\ 2 \end{pmatrix}$ For C, the image $C' = \begin{pmatrix} 0 \\ 0 \end{pmatrix} + \begin{pmatrix} -2 \\ -3 \end{pmatrix}$ $= \begin{pmatrix} -2 \\ -3 \end{pmatrix}$ The coordinate of the vertices of image triangle $A'B'C'$ are $A'(1, 1),$ $B'(-3, 2)$ and $C'(-2, -3)$

Reflection and its properties

To reflect an object in a given line, for example the x-axis, use the x-axis as a mirror line and use one of the characteristics of a plane mirror, which is “the distance between the object and the mirror is the same as the distance between the mirror and the image”. For example, in the diagram below we have a reflection in the x-axis. The x-axis is used as a mirror. The perpendicular distance between the object $P(2, 3)$ and the mirror is 3 up, counting 3 down from the mirror line gives us $P'(2, -3)$

Reflection in the axis Reflection in the axis [Figure] [Figure] Given T (1,0) $\xrightarrow{x\text{-axis}}$ (1, 0) Given T (1,0) $\xrightarrow{y\text{-axis}}$ (-1, 0) Also, T (0,1) $\xrightarrow{x\text{-axis}}$ (0, -1) Also, T (0,1) $\xrightarrow{y\text{-axis}}$ (0, 1) If M is the matrix of the If M is the matrix of the transformation, transformation, $M =$ $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ $M =$ $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ To find the image of an object, multiply the transformation matrix, M, by the object $\begin{pmatrix} x \\ y \end{pmatrix}.$ The table below summarises the various reflections and how to get the image.

Type of reflection Point Matrix to Image multiply by $x$ -axis ( $y$ = 0) $(x, y)$ $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ $y$ -axis ( $x$ = 0) $(x, y)$ $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ $y = x$ $(x, y)$ $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$ $y = -x$ $(x, y)$ $\begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix}$ $\begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$