HSC Maths and Stats 1 Imp Questions 2025 PDF
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Shree L. R. Tiwari College
2025
Akash Agrawal
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This document contains important questions for HSC Class 12th Commerce Maths & Stats 1, including concepts such as mathematical logic, differentiation, and integration. Board paper solutions and video lectures are available. The document also covers quantitative aptitude, with exercises and additional question banks for practice.
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MATHS & STATS 1 MOST IMPORTANT QUESTIONS HSC Class 12th Commerce #KarnaHiHain #Ron...
MATHS & STATS 1 MOST IMPORTANT QUESTIONS HSC Class 12th Commerce #KarnaHiHain #RoneKaNahiFhodneKa #TheMathsTutor #TheMathsTutor R AW AL ash A G ak E M ath s tu tor TH AkashAgrawalTheMathsTutor AkashAgrawalTheMathsTutor Tagore Nagar, Vikhroli - E @akashpagrawal @akashp.agrawal Mumbai Scan the following QR Codes to watch the Video Lectures of the following chapters: ONE SHOT LECTURES OF ALL tOPICS OF MATHS 1 & MATHS 2 Complete chapter INTEGRation Differential probability differentiation equation distribution linear mathematical application of regression logic time series derivatives complete miscellaneous objectives maths 1 all chapter maths 2 all chapter objectives objectives Past board paper solutions (5 papers) :- Sept 2021 to July 2023 OF MATHS 1 & MATHS 2 maths 1 board maths 2 board paper solution paper solution Note:- board paper solutions for March 2024 & july 2024 will be uploaded before board exam Quantitative Aptitude (Maths, Stats & LR) June 25/Sept 25 Nov 22/May 120 hrs All concepts & basics, Module Exercises, Additional Question Banks, Chapterwise PYQs of previous 13 years & MTPs Dedicated - Chapterwise, sectionwise, #TheMathsTutor complete syllabus test on OMR sheets Extra support for non-Maths background students For details :- 8690369038 Connect with me on social media platforms AkashAgrawalTheMathsTutor AkashAgrawalTheMathsTutor Tagore Nagar, Vikhroli - E @akashpagrawal @akashp.agrawal Mumbai INDEX Sr. Page Chapter No. No. 1 Mathematical Logic 1 2 Matrices 35 3 Differentiation 89 4 Applications of Derivatives 103 5 Integration 116 6 Definite Integration 141 7 Application of Definite Integration 152 8 Differential Equation and Applications 160 Answers 174 1 Mathematical logic For example: Let's Study i) 2 is a prime number. Statement ii) Every rectangle is a square. Logical connectives iii) The Sun rises in the West. Quantifiers and quantified statements iv) Mumbai is the capital of Maharashtra. Statement patterns and logical Truth value of a statement: equivalence A statement is either true or false. The Algebra of statements truth value of a ‘true’ statement is denoted by T (TRUE) and that of a false statement is denoted Venn diagrams by F (FALSE). Introduction: Example 1: Observe the following sentences. Mathematics is an exact science. Every i) The Sun rises in the East. statement must be precise. There has to be ii) The square of a real number is negative. proper reasoning in every mathematical proof. Proper reasoning involves Logic. Logic related iii) Sum of two odd numbers is odd. to mathematics has been developed over last 100 iv) Sum of opposite angles in a cyclic rectangle years or so. The axiomatic approach to logic is 180°. was first propounded by the English philosopher Here, the truth value of statements (i) and and mathematician George Boole. Hence it is (iv) is T, and that of (ii) and (iii) is F. known as Boolean logic or mathematical logic Note: The sentences like exclamatory, or symbolic logic. interrogative, imperative are not considered as statements. The word ‘logic’ is derived from the Greek word ‘Logos’ which means reason. Thus Logic Example 2: Observe the following sentences. deals with the method of reasoning. Aristotle i) May God bless you! (382-322 B.C.), the great philosopher and ii) Why are you so unhappy? thinker laid down the foundations of study of logic in a systematic form. The study of logic iii) Remember me when we are parted. helps in increasing one’s ability of systematic iv) Don’t ever touch my phone. and logical reasoning and develop the skill of v) I hate you! understanding validity of statements. vi) Where do you want to go today? 1.1 Statement: The above sentences cannot be assigned A statement is a declarative sentence which truth values, so none of them is a statement. is either true or false but not both simultaneously. The sentences (i) and (v) are exclamatory. Statements are denoted by letters like p,q,r,.... The sentences (ii) and (vi) are interrogative. The sentences (iii) and (iv) are imperative. #AkashAgrawalTheMathsTutor 1 ii) In this sentence ‘or’ is used for indicating EXERCISE 1.2 approximate number of students and not Ex. 1: Express the following statements in as a connective. Therefore, it is a simple symbolic form. statement and it is expressed as i) e is a vowel or 2 + 3 = 5 p : Four or five students did not attend the lectures. ii) Mango is a fruit but potato is a vegetable. iii) Milk is white or grass is green. Ex. 2: Write the truth values of the following iv) I like playing but not singing. statements. v) Even though it is cloudy, it is still raining. i) India is a democratic country or China is a communist country. Ex. 2: Write the truth values of following ii) India is a democratic country or China is statements. not a communist country. i) Earth is a planet and Moon is a star. iii) India is not a democratic country or China ii) 16 is an even number and 8 is a perfect is a communist country. square. iv) India is not a democratic country or China iii) A quadratic equation has two distinct roots is not a communist country. or 6 has three prime factors. iv) The Himalayas are the highest mountains Solution: p : India is a democratic country. but they are part of India in the North East. q : China is a communist country. C) Negation (∼): p is true; q is true. The denial of an assertion contained in a i) True (T), since both the sub-statements are statement is called its negation. true i.e. both “India is a democratic country” The negation of a statement is generally and “China is a communist country” are formed by inserting the word “not” at some true. (As T ∨ T = T) proper place in the statement or by prefixing the ii) True (T), since first sub-statements “India statement with “it is not the case that” or “it is is a democratic country” is true and second false that” or “it is not true that”. sub-statement “China is not a communist The negation of a statement p is written as country” is false. (As T ∨ F = T) ∼ p (read as “negation p” or “not p”) in symbolic iii) True (T), since first sub-statements “India form. is not a democratic country” is false For example: and second sub-statement “China is a communist country” is true. (As F ∨ T = T) Let p : 2 is an even number ∼ p : 2 is not an even number. iv) False (F), since both the sub-statements “India is not a democratic country” and or ∼ p : It is not the case that 2 is an even “China is not a communist country” are number false. (As F ∨ F = F) or ∼ p : It is false that 2 is an even number #AkashAgrawalTheMathsTutor 6 The truth table of negation (∼ p) D) Conditional statement (Implication, →) Table 1.4 If two simple statements p and q are p ∼p connected by the group of words “If... then...”, then the resulting compound statement “If p then T F q” is called a conditional statement (implication) F T and is written in symbolic form as “p → q” (read Note: Negation of a statement is equivalent to as “p implies q”). the complement of a set in set theory. For example: SOLVED EXAMPLES i) Let p : There is rain q : The match will be cancelled Ex. 1: Write the negation of the following statements. then, p → q : If there is rain then the match will be cancelled. i) p : He is honest. ii) Let p : r is a rational number. ii) q : p is an irrational number. q : r is a real number. Solution: then, p → q : If r is a rational number then r i) ∼ p : He is not honest is a real number. or ∼ p : It is not the case that he is honest The truth table for conditional statement or ∼ p : It is false that he is honest. (p → q) ii) ∼ q : p is not an irrational number. Table 1.5 or ∼ q : p q p→q or ∼ q : T T T T F F F T T EXERCISE 1.3 F F T 1. Write the negation of each of the following statements. SOLVED EXAMPLES i) All men are animals. ii) − 3 is a natural number. Ex. 1: Express the following statements in the iii) It is false that Nagpur is capital of symbolic form. Maharashtra i) If the train reaches on time, then I can catch iv) 2 + 3 ≠ 5 the connecting flight. 2. Write the truth value of the negation of each ii) If price increases then demand falls. of the following statements. Solution: i) 5 is an irrational number i) Let p : The train reaches on time ii) London is in England q : I can catch the connecting flight. iii) For every x ∈ N , x + 3 < 8. Therefore, p → q is symbolic form. #AkashAgrawalTheMathsTutor 7 Therefore, p ↔ (∼ q) is the symbolic form. Ex. 3: If p and q are true and r and s are false, Given statement, p is false F. find the truth value of each of the following. q is true T. i) (p ↔ ∼ q) ∧ (r ↔ ∼ s) ∴ ∼ q is false F. ii) (p → r) ∨ (q → s) ∴ p ↔ (∼ q) ≡ F ↔ F ≡ T iii) ∼ [(p ∧ ∼ s) ∨ (q ∧ ∼ r)] ∴ the truth value of given statement is T. Solution: iii) Let p : i is a real number. i) Without truth table : (p ↔ ∼ q) ∧ (r ↔ ∼ s) ∴ ∼ p : It is not true that i is a real number. = (T ↔ ∼ T) ∧ (F ↔ ∼ F) Therefore, ∼ p : is the symbolic form. = (T ↔ F) ∧ (F ↔ T) p is false F. =F∧ F =F ∴ ∼ p : is true T. ∴ the truth value of the given statement is T. Construct truth table Table 1.7 p q r s ∼q ∼s p↔∼q r↔∼s (p ↔ ∼ q) ∧ (r ↔ ∼ s) T T F F F T F F F ii) Without truth table : Construct truth table (p → r) ∨ (q → s) = (T → F) ∨ (T → F) Table 1.9 =F∨F (Note: Construct truth table and complete =F your solution) Construct truth table EXERCISE 1.4 Table 1.8 Ex. 1: Write the following statements in p q r s p → r q → s (p → r) ∨ (q → s) symbolic form. T T F F F F F i) If triangle is equilateral then it is equiangular. ii) Without truth table : (Activity) ii) It is not true that “i” is a real number. ∼ [(p ∧ ∼ s) ∨ (q ∧ ∼ r)] iii) Even though it is not cloudy, it is still raining. = ∼ [( ∧∼ )∨( ∧ )] iv) Milk is white if and only if the sky is not = ∼ [( ∧ T) ∨ (T ∧ T)] blue. =∼( ∨ ) v) Stock prices are high if and only if stocks are rising. =∼( ) vi) If Kutub-Minar is in Delhi then Taj-Mahal = is in Agra. #AkashAgrawalTheMathsTutor 10 Ex. 2: Find truth value of each of the following 1.2.1 Quantifiers and Quantified statements: statements. i) For every x ∈ R, x2 is non negative. We i) It is not true that 3 − 7i is a real number. shall now see how to write this statement ii) If a joint venture is a temporary partnership, using symbols. ‘∀x’ is used to denote “For then discount on purchase is credited to the all x”. supplier. Thus, the above statement may be written iii) Every accountant is free to apply his own in mathematical notation " z ∈ R, z2 ≥ 0. accounting rules if and only if machinery is The symbol ‘"’ stands for “For all values an asset. of”. This is known as universal quantifier. iv) Neither 27 is a prime number nor divisible by 4. ii) Also we can get x ∈ N such that x + 4 = 7. To write this in symbols we use the symbol v) 3 is a prime number and an odd number. ∃ x to denote “there exists x”. Thus, we have ∃ x ∈ N such that x + 4 = 7. Ex. 3: If p and q are true and r and s are false, find the truth value of each of the following The symbol ∃ stands for “there exists”. This compound statements. symbol is known as existential quantifier. i) p ∧ (q ∧ r) Thus, there are two types of quantifiers. ii) (p → q) ∨ (r ∧ s) a) Universal quantifier (") iii) ∼ [(∼ p ∨ s) ∧ (∼ q ∧ r)] b) Existential quantifier (∃) iv) (p → q) ↔ ∼ (p ∨ q) Quantified statement: v) [(p ∨ s) → r] ∨ ∼ [∼ (p → q) ∨ s] An open sentence with a quantifier becomes vi) ∼ [p ∨ (r ∧ s)] ∧ ∼ [(r ∧ ∼ s) ∧ q] a statement and is called a quantified statement. Ex. 4: Assuming that the following statements are true, SOLVED EXAMPLES p : Sunday is holiday, q : Ram does not study on holiday, Ex. 1: Use quantifiers to convert each of the find the truth values of the following following open sentences defined on N, into a statements. true statement. i) Sunday is not holiday or Ram studies on i) 2x + 3 = 11 holiday. ii) If Sunday is not holiday then Ram studies ii) x3 < 64 on holiday. iii) x + 5 < 9 iii) Sunday is a holiday and Ram studies on holiday. Solution: Ex. 5: If p : He swims i) ∃ x ∈ N such that 2x + 3 = 11. It is a true q : Water is warm statement, since x = 4 ∈ N satisfies 2x + 3 = 11. Give the verbal statements for the following symbolic statements. ii) x3 < 64 ∃ x ∈ N such that it is a true statement, since x = 1 or 2 or 3 ∈ N satisfies i) p↔∼q x3 < 64. ii) ∼ (p ∨ q) iii) ∃ x ∈ N such that x + 5 < 9. It is a true iii) q → p statement for x = 1 or 2 or 3 ∈ N satisfies iv) q ∧ ∼ p x + 5 < 9. #AkashAgrawalTheMathsTutor 11 Ex. 2: If A = {1, 3, 5, 7} determine the truth For example: value of each of the following statements. i) (p ∨ q) → r i) ∃ x ∈ A, such that x2 < 1. ii) ∃ x ∈ A, such that x + 5 ≤ 10 ii) p ∧ (q ∧ r) iii) " x ∈ A, x + 3 < 9 iii) ∼ (p ∨ q) are statement patterns Solution: Note: While preparing truth tables of the given statement patterns, the following points should i) No number in set A satisfies x2 < 1, since be noted. the square of every natural number is 1 or greater than 1. i) If a statement pattern involves n component ∴ the given statement is false, hence its statements p, q, r,..... and each of p, q, r,..... truth value is F. has 2 possible truth values namely T and F, then the truth table of the statement pattern ii) Clearly, x = 1, 3 or 5 satisfies x + 5 ≤ 10. consists of 2n rows. So the given statement is true, hence truth value is T. ii) If a statement pattern contains “m” iii) Since x = 7 ∈ A does not satisfy x + 3 < 9, connectives and “n” component statements the given statement is false. Hence its truth then the truth table of the statement pattern value is F. consists of (m + n) columns. iii) Parentheses must be introduced whenever EXERCISE 1.5 necessary. Ex. 1: Use quantifiers to convert each of the For example: following open sentences defined on N, into a ∼ (p → q) and ∼ p → q are not the same. true statement. i) x2 + 3x − 10 = 0 SOLVED EXAMPLES ii) 3x − 4 < 9 iii) n2 ≥ 1 Ex. 1: Prepare the truth table for each of the iv) 2n − 1 = 5 following statement patterns. v) Y+4>6 i) [(p → q) ∨ p] → p vi) 3y − 2 ≤ 9 ii) ∼ [p ∨ q] → ∼ (p ∧ q) Ex. 2: If B = {2, 3, 5, 6, 7} determine the truth iii) (∼ p ∨ q) ∨ ∼ q value of each of the following. iv) (p ∧ r) ∨ (q ∧ r) i) " x ∈ B such that x is prime number. ii) ∃ n ∈ B, such that n + 6 > 12 Solution: iii) ∃ n ∈ B, such that 2n + 2 < 4 i) [(p → q) ∨ p) → p iv) " y ∈ B such that y2 is negative v) " y ∈ B such that (y − 5) ∈ N Truth table 1.10 p q p → q (p → q) ∨ p [(p → q) ∨ p] 1.3 Statement Patterns and Logical →p Equivalence: T T T T T A) Statement Patterns: T F F T T Let p, q, r,..... be simple statements. F T T T F A compound statement obtained from these F F T T F simple statements by using one or more of the connectives ∧, ∨, →, ↔, is called a statement ii) ∼ (p ∨ q) → ∼ (p ∧ q) pattern. 12 #AkashAgrawalTheMathsTutor Truth table 1.23 i) q ∨ [∼ (p ∧ q)] p q ∼q ∼q∧p (∼ q ∧ p) ∧ q ii) (∼ q ∧ p) ∧ (p ∧ ∼ p) T T F F F iii) (p ∧ ∼ q) → (∼ p ∧ ∼ q) T F T T F iv) ∼ p → (p → ∼ q) F T F F F 3. Prove that each of the following statement F F T F F pattern is a tautology. In the table 1.23, all the entries in the last i) (p ∧ q) → q column are F. ii) (p → q) ↔ (∼ q → ∼ p) Therefore, the given statement pattern is a contradiction. iii) (∼ p ∧ ∼ q) → (p → q) iv) The truth table for p → (∼ q ∨ r) iv) (∼ p ∨ ∼ q) ↔ ∼ (p ∧ q) Truth table 1.24 4. Prove that each of the following statement p q r ∼q ∼q∨r p → (∼ q ∨ r) pattern is a contradiction. T T T F T T i) (p ∨ q) ∧ (∼ p ∧ ∼ q) T T F F F F ii) (p ∧ q) ∧ ∼ p T F T T T T iii) (p ∧ q) ∧ (∼ p ∨ ∼ q) T F F T T T iv) (p → q) ∧ (p ∧ ∼ q) F T T F T T 5. Show that each of the following statement F T F F F T pattern is a contingency. F F T T T T i) (p ∧ ∼ q) → (∼ p ∧ ∼ q) F F F T T T ii) (p → q) ↔ (∼ p ∨ q) In the table 1.24, the entries in the last iii) p ∧ [(p → ∼ q) → q] column are neither all T nor all F. iv) (p → q) ∧ (p → r) Therefore, the given statement pattern is a contingency. 6. Using the truth table, verify i) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) EXERCISE 1.6 ii) p → (p → q) ≡ ∼ q → (p → q) 1. Prepare truth tables for the following iii) ∼ (p → ∼ q) ≡ p ∧ ∼ (∼ q) ≡ p ∧ q statement patterns. iv) ∼ (p ∨ q) ∨ (∼ p ∧ q) ≡ ∼ p i) p → (∼ p ∨ q) ii) (∼ p ∨ q) ∧ (∼ p ∨ ∼ q) 7. Prove that the following pairs of statement patterns are equivalent. iii) (p ∧ r) → (p ∨ ∼ q) i) p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r) iv) (p ∧ q) ∨ ∼ r ii) p ↔ q and (p → q) ∧ (q → p) 2. Examine whether each of the following iii) p → q and ∼ q → ∼ p and ∼ p ∨ q statement patterns is a tautology, a contradiction or a contingency iv) ∼ (p ∧ q) and ∼ p ∨ ∼ q. #AkashAgrawalTheMathsTutor 16 E) Negation of a compound statement: 3) Negation of negation: We have studied the negation of simple Let p be a simple statement. statements. Negation of a simple statement is Truth table 1.25 obtained by inserting “not” at the appropriate p ∼p ∼ (∼ p) place in the statement e.g. the negation of “Ram is tall” is “Ram is not tall”. But writing negations T F T of compound statements involving conjunction., F T F disjunction, conditional, biconditional etc. is not straight forward. From the truth table 1.25, we see that ∼ (∼ p) ≡ p 1) Negation of conjunction: Thus, the negation of negation of a In section 1.3(B) we have seen that ∼ (p statement is the original statement - ∼ (∼ p) ≡ p. ∧ q) ≡ ∼ p ∨ ∼ q. It means that negation of the conjunction of two simple statements is the For example: disjunction of their negation. Let p : 5 is an irrational number. Consider the following conjunction. The negation of p is given by “Parth plays cricket and chess.” ∼ p : 5 is not an irrational number. Let p : Parth plays cricket. ∼ (∼ p) : 5 is an irrational number. q : Parth plays chess. Therefore, negation of negation of p is Given statement is p ∧ q. ∼ (∼ p) i.e. it is not the case that 5 is not an You know that ∼ (p ∧ q) ≡ ∼ p ∨ ∼ q irrational number. ∴ negation is Parth doesn’t play cricket or OR it is false that 5 is not an irrational he doesn’t play chess. number. OR 5 is an irrational number. 2) Negation of disjunction: In section 1.3(B) we have seen that 4) Negation of Conditional (Implication): ∼ (p ∨ q) ≡ ∼ p ∧ ∼ q. It means that negation of You know that p → q ≡ ∼ p ∨ q the disjunction of two simple statements is the ∴ ∼ (p → q) ≡ ∼ (∼ p ∨ q) conjunction of their negation. ≡ ∼ (∼ p) ∧ ∼ q... by De-Morgan's law For ex: The number 2 is an even number or the number 2 is a prime number. ∴ ∼ (p → q) ≡ p ∧ ∼ q Let p : The number 2 is an even number. We can also prove this result by truth table. q : The number 2 is a prime number. Truth table 1.26 ∴ given statement : p ∨ q. p q p → q ∼ (p → q) ∼q p∧∼q You know that ∼ (p ∨ q) ≡ ∼ p ∧ ∼ q 1 2 3 4 5 6 ∴ negation is “The number 2 is not an T T T F F F even number and the number 2 is not a prime T F F T T T number”. F T T F F F F F T F F F #AkashAgrawalTheMathsTutor 18 All the entries in the columns 4 and 6 of 5) Negation of Biconditional (Double table 1.26 are identical. implication): ∴ ∼ (p → q) ≡ p ∧ ∼ q Consider the biconditional p ↔ q. e.g. If every planet moves around the Sun Method 1: then every Moon of the planet moves around the We have seen that Sun. (p ↔ q) ≡ (p → q) ∧ (q → p) Negation of the given statement is, Every ∴ ∼ (p ↔ q) ≡ ∼ [(p → q) ∧ (q → p)] planet moves around the Sun but (and) every ≡ ∼ (p → q) ∨ ∼ (q → p) Moon of the planet does not move around the... by De-Morgans law Sun. ≡ (p ∧ ∼ q) ∨ (q ∧ ∼ p)... by negation of the conditional statement ∴ ∼ (p ↔ q) ≡ (p ∧ ∼ q) ∨ (q ∧ ∼ p) Method 2: We also prove this by using truth table 1.27. Truth Table 1.27 p q p ↔ q ∼ (p ↔ ∼ q) ∼p ∼q p ∧ ∼ q q ∧ ∼ p (p ∧ ∼ q) ∨ (q ∧ ∼ p) 1 2 3 4 5 6 7 8 9 T T T F F F F F F T F F T F T T F T F T F T T F F T T F F T F T T F F F Since all the entries in the columns 4 and 9 of truth table 1.27 are identical. ∴ ∼ (p ↔ q) ≡ (p ∧ ∼ q) ∨ (q ∧ ∼ p). For example: 2n is divisible by 4 if and only if For example: Consider the statement pattern n is an even integer. (∼ p ∧ q) ∨ (p ∨ ∼ q). Its negation is given by : Let p : 2n is divisible by 4 i.e. ∼ [(∼ p ∧ q) ∨ (p ∨ ∼ q)] q : n is an even integer. ≡ (p ∨ ∼ q) ∧ (∼ p ∧ q) 6) Negation of a quantified statement: Therefore, negation of the given statement is “2n is divisible by 4 and n is not an even integer While forming negation of a quantified or n is an even integer and 2n is not divisible statement, we replace the word ‘all’ by ‘some’, by 4”. “for every” by “there exists” and vice versa. Note: SOLVED EXAMPLES Negation of a statement pattern involving Ex. 1: Write negation of each of the following one or more of the simple statements p, q, r,... statements : and one or more of the three connectives ∧, ∨, ∼ can be obtained by replacing ∧ by ∨, ∨ by ∧ and i) All girls are sincere replaicng p, q, r....by ∼p, ∼q, ∼r..... #AkashAgrawalTheMathsTutor 19 ii) If India is playing world cup and Rohit is Converse: q → p i.e. If a man is happy the captain, then we are sure to win. then he is rich. iii) Some bureaucrats are efficient. Inverse: ∼ p → ∼ q i.e. If a man is not rich then he is not happy. Solution: Contrapositive: ∼ q → ∼ p i.e. If a man is i) The negation is, not happy then he is not rich. Some girls are not sincere ii) Let p : The train reaches on time. OR, There exists a girl, who is not sincere. q : I can catch the connecting flight. ii) Let p : India is playing world cup Therefore, the symbolic form of the given q : Rohit is the captain statement is p → q. r : We win. Converse, q → p i.e. The given compound statement is Inverse i.e. (p ∧ q) → r Contrapositive i.e. Therefore, the negation is, Ex. 3: Using the rules of negation, write the ∼ [(p ∧ q) → r] ≡ (p ∧ q) ∧ ∼ r negation of the following : India is playing world cup and Rohit is the i) (∼ p ∧ r) ∨ (p ∨ ∼ r) captain and we are not sure to win. ii) (p ∨ ∼ r) ∧ ∼ q iii) The negation is, all bureaucrats are not iii) The crop will be destroyed if there is a efficient. flood. Converse, Inverse and contrapositive: Solution: Let p and q be simple statements and i) The negation of (∼ p ∧ r) ∨ (p ∨ ∼ r) is let p → q be the implication of p and q. ∼ [(∼ p ∧ r) ∨ (p ∨ ∼ r)] Then, i) The converse of p → q is q → p. ≡ ∼ (∼ p ∧ r) ∧ ∼ (p ∨ ∼ r) ii) Inverse of p → q is ∼ p → ∼ q.... by De-Morgan's law iii) Contrapositive of p → q is ∼ q → ∼ p. ≡ (p ∨ ∼ r) ∧ (∼ p ∧ r) For example: Write the converse, inverse and... by De-Morgan's law and contrapositive of the following compound ∼ (∼ p) ≡ p and ∼ (∼ r) = r. statements. ii) The negation of (p ∨ ∼ r) ∧ ∼ q is i) If a man is rich then he is happy. ii) If the train reaches on time then I can catch ∼ [(p ∨ ∼ r) ∧ ∼ q] the connecting flight. ≡ ∼ (p ∨ ∼ r) ∨ ∼ (∼ q)... by De Morgan's law Solution: ≡ (∼ p ∧ r) ∨ q i) Let p : A man is rich. q : He is happy.... by De Morgan's law and ∼ (∼ q) ≡ q. Therefore, the symbolic form of the given statement is p → q. #AkashAgrawalTheMathsTutor 20 iii) Let p : The crop will be destroyed. 2. Using the rules of negation, write the q : There is a flood. negations of the following : Therefore, the given statement is q → p i) (p → r) ∧ q and its negation is ∼ (q → p) ≡ q ∧ ∼ p ii) ∼ (p ∨ q) → r i.e. the crop will not be destroyed and there iii) (∼ p ∧ q) ∧ (∼ q ∨ ∼ r) is a flood. 3. Write the converse, inverse and contrapositive of the following statements. EXERCISE 1.8 i) If it snows, then they do not drive the 1. Write negation of each of the following car. statements. ii) If he studies, then he will go to college. i) All the stars are shining if it is night. ii) ∀ n ∈ N, n + 1 > 0 4. With proper justification, state the negation of each of the following. iii) ∃ n ∈ N, (n2 + 2) is odd number i) (p → q) ∨ (p → r) iv) Some continuous functions are differentiable. ii) (p ↔ q) ∨ (∼ q → ∼ r) iii) (p → q) ∧ r 1.4 Algebra of statements: The statement patterns, under the relation of logical equivalence, satisfy various laws. We have already proved a majority of them and the rest are obvious. Now, we list these laws for ready reference. 1. p∨p≡p Idempotent laws p∧p≡p 2. p ∨ (q ∨ r) Associative laws p ∧ (q ∧ r) ≡ (p ∨ q) ∨ r ≡ (p ∧ q) ∧ r ≡p∨q∨r ≡ (p ∧ q) ∧ r 3. (p ∨ q) ≡ q ∨ p Commutative laws p∧q≡q∧p 4. p ∨ (q ∧ r) Distributive laws p ∧ (q ∨ r) ≡ (p ∨ q) ∧ (p ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) 5. p∨c≡p Identity laws p∧c≡c p∨t≡t p∧t≡p 6. p∨∼p≡t Complement laws p∧∼p≡c ∼t≡c ∼c≡t 7. ∼ (∼ p) ≡ p Involution law (law of double negation) 8. ∼ (p ∨ q) DeMorgan’s laws ∼ (p ∧ q) ≡∼p∧∼q ≡∼p∨∼q 9. p→q Contrapositive law ≡∼q→∼p #AkashAgrawalTheMathsTutor 21 Note: In case of three simple statements p,q,r, ≡ [(p ∧ ∼ p) ∨ (q ∧ ∼ p)] → q we note the following :... by Distributive law i) p ∧ q ∧ r is true if and only if p, q, r are all ≡ [(c ∨ (q ∧ ∼ p)] → q true and p ∧ q ∧ r is false even if any one of p, q, r is false.... by Complement law ii) p ∨ q ∨ r is false if and only if p, q, r are all ≡ (∼ p ∧ q) → q... by Commutative law false, otherwise it is true. ≡ ∼ (∼ p ∧ q) ∨ q... by ∼ ∼ (p → q) ≡ ∼ (p ∧ ∼ q) ≡ ∼ p ∨ q SOLVED EXAMPLES Ex. 1: Without using truth table, show that ≡ [(p ∨ ∼ q) ∨ q... by De Morgan's law i) p ∨ (q ∧ ∼ q) ≡ p ≡ p ∨ (∼ q ∨ q)]... by Associative law ii) ∼ (p ∨ q) ∨ (∼ p ∧ q) ≡ ∼ p ≡p∨t iii) p ∨ (∼ p ∧ q) ≡ p ∨ q ≡t Solution: EXERCISE 1.9 i) p ∨ (q ∧ ∼ q) ≡p∨c... by complement law 1. Without using truth table, show that ≡p... by Identity law i) p ↔ q ≡ (p ∧ q) ∨ (∼ p ∧ ∼ q) ii) ∼ (p ∨ q) ∨ (∼ p ∧ q) ii) p ∧ [(∼ p ∨ q) ∨ ∼ q] ≡ p ≡ (∼ p ∧ ∼ q) ∨ (∼ p ∧ q) iii) ∼ [(p ∧ q) → ∼ q] ≡ p ∧ q... by De Morgans law iv) ∼ r → ∼ (p ∧ q) ≡ [∼ (q → r)] → ∼ p v) (p ∨ q) → r ≡ (p → r) ∧ (q → r) ≡ ∼ p ∧ (∼ q ∨ q)... by Distributive law 2. Using the algebra of statement, prove that ≡∼p∧t i) [p ∧ (q ∨ r)] ∨ [∼ r ∧ ∼ q ∧ p] ≡ p... by Complement law ii) (p ∧ q) ∨ (p ∧ ∼ q) ∨ (∼ p ∧ ∼ q) ≡ ≡∼p... by Identity law p ∨ ∼ q) iii) p ∨ (∼ p ∧ q) iii) (p ∨ q) ∧ (∼ p ∨ ∼ q) ≡ (p ∨ ∼ q) ∧ (∼ p ∨ q) ≡ (p ∨ ∼ p) ∧ (p ∨ q)... by Distributive law 1.5 Venn Diagrams: ≡ t ∧ (p ∨ q)... by Complement law We have already studied Venn Diagrams while studying set theory. Now we try to ≡p∨q... by Identity law investigate the similarly between rules of logical connectives and those of various operations on Ex. 2: Without using truth table, prove that sets. [(p ∨ q) ∧ ∼ p] → q is a tautology. The rules of logic and rules of set theory go Solution: hand in hand. [(p ∨ q) ∧ ∼ p] → q #AkashAgrawalTheMathsTutor 22 The Venn diagram (fig. 1.17) represents the By Venn diagrams (fig. 1.19), we observe truth of the given statement. that truth set of statements (i) and (ii) are equal. Hence, statements (i) and (ii) are logically Ex. 2: Draw the Venn diagram for the truth of equivalent. the following statements. i) There are students who are not scholars. EXERCISE 1.10 ii) There are scholars who are students. 1. Represent the truth of each of the following iii) There are persons who are students and statements by Venn diagrams. scholars. i) Some hardworking students are obedient. Solution: ii) No circles are polygons. Let us choose the universal set. iii) All teachers are scholars and scholars U : The set of all human beings. are teachers. Let S : The set of all scholars. iv) If a quadrilateral is a rhombus, then it T : The set of all students. is a parallelogram. i) ii) iii) 2. Draw a Venn diagram for the truth of each of the following statements. i) Some sharebrokers are chartered accountants. ii) No wicket keeper is bowler, in a cricket Fig. 1.18 team. We observe that (by Venn diagram) truth 3. Represent the following statements by Venn set of statements (ii) and (iii) are equal. Hence, diagrams. statements (ii) and (iii) are logically equivalent. i) Some non resident Indians are not rich. ii) No circle is rectangle. Ex. 3: Using the Venn diagram, examine the logical equivalence of the following statements. iii) If n is a prime number and n ≠ 2, then it is odd. i) Some politicians are actors. ii) There are politicians who are actors. Let's Remember iii) There are politicians who are not actors. 1. Statement: Declarative sentence which Solution: is either true or false, but not both Let us choose the universal set. symultaneously. U : The set of all human beings. ∗ Imperative, exclamatory, interrogative Let P : The set of all politicians. and open sentences are not statements. A : The set of all actors. ∗ The symbol ‘∀’ stands for “all values of”. It is universal quantifier. i) ii) iii) ∗ The symbol ‘∃’ stands for “there exists”. It is known as existential quantifier. ∗ An open sentence with a quantifier Fig. 1.19 becomes a quantified statement. #AkashAgrawalTheMathsTutor 27 iv) Kanchanganga is in India and Everest 16. State the dual of each of the following is in Nepal. statements by applying the principle of v) If x∈A∩B, then x∈A and x∈B. duality. i) (p ∧ ∼ q) ∨ (∼ p ∧ q) ≡ (p ∨ q) ∧∼ (p ∧ q) 11. Construct the truth table for each of the ii) p ∨ (q ∨ r) ≡ ∼ [(p ∧ q) ∨ (r ∨ s)] following statement pattern. iii) 2 is even number or 9 is a perfect i) (p ∧ ∼ q) ↔ (q → p) square. ii) (∼ p ∨ q) ∧ (∼ p ∧ ∼ q) iii) (p ∧ r) → (p ∨ ∼ q) 17. Rewrite the following statements without using the connective ‘If... then’. iv) (p ∨ r) → ∼ (q ∧ r) i) If a quadrilateral is rhombus then it is v) (p ∨ ∼ q) → (r ∧ p) not a square. 12. What is tautology? What is contradiction? ii) If 10 − 3 = 7 then 10 × 3 ≠ 30. Show that the negation of a tautology iii) If it rains then the principal declares a is a contradiction and the negation of a holiday. contradiction is a tautology. 18. Write the dual of each of the following. 13. Determine whether following statement i) (∼ p ∧ q) ∨ (p ∧ ∼ q) ∨ (∼ p ∧ ∼ q) pattern is a tautology, contradiction, or contingency. ii) (p ∧ q) ∧ r ≡ p ∧ (q ∧ r) i) [(p ∧ q) ∨ (∼ p)] ∨ [p ∧ (∼ q)] iii) p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (q ∨ r) ii) [(∼ p ∧ q) ∧ (q ∧ r)] ∨ (∼ q) iv) ∼ (p ∨ q) ≡ ∼ p ∧ ∼ q iii) [∼ (p ∨ q) → p] ↔ [(∼ p) ∧ (∼ q)] 19. Consider the following statements. iv) [∼ (p ∧ q) → p] ↔ [(∼ p) ∧ (∼ q)] i) If D is dog, then D is very good. v) [p → (∼ q ∨ r)] ↔ ∼ [p → (q → r)] ii) If D is very good, then D is dog. 14. Using the truth table, prove the following iii) If D is not very good, then D is not a logical equivalences. dog. i) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) iv) If D is not a dog, then D is not very good. ii) [∼ (p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r Identify the pairs of statements having the iii) p ∧ (∼ p ∨ q) ≡ p ∧ q same meaning. Justify. iv) p ↔ q ≡ ∼ (p ∧ ∼ q) ∧ ∼ (q ∧ ∼ p) v) ∼ p ∧ q ≡ (p ∨ q)] ∧ ∼p 20. Express the truth of each of the following statements by Venn diagrams. 15. Write the converse, inverse, contrapositive i) All men are mortal. of the following statements. ii) Some persons are not politician. i) If 2 + 5 = 10, then 4 + 10 = 20. iii) Some members of the present Indian ii) If a man is bachelor, then he is happy. cricket are not committed. iii) If I do not work hard, then I do not iv) No child is an adult. prosper. #AkashAgrawalTheMathsTutor 33 2 Matrices Let's Study Types of Matrices Algebra of Matrices Properties of Matrices Elementary Transformation Inverse of Matrix Application of Matrices Determinant of a Matrix Let's Recall Determinant of a Matrix 2.1 Introduction: The theory of matrices was developed by the mathematician Arthur Cayley. Matrices are useful in expressing numerical information in a compact form. They are effectively used in expressing different operations. Hence they are essential in economics, finance, business and statistics. Definition: A rectangular arrangement of mn numbers in m rows and n columns, enclosed in [ ] or () is called a matrix of order m by n. A matrix by itself does not have a value or any special meaning. The order of a matrix is denoted by m × n, read as m by n. Each member of a matrix is called an element of the matrix. Matrices are generally denoted by capital letters like A, B, C, …. and their elements are denoted by small letters like aij, bij, cij, ….. etc. where aij is the element in ith row and jth column of the matrix A. 2 3 9 For example: i) A 1 0 7 here a32 = −2 4 2 1 A is a matrix having 3 rows and 3 columns. The order of A is 3×3. There are 9 elements in the matrix A. 1 5 ii) B 2 6 0 9 B is a matrix having 3 rows and 2 columns. The order of B is 3×2. There are 6 elements in the matrix B. In general, a matrix of order m × n is represented by #AkashAgrawalTheMathsTutor 35 5 5 30 6 9 3 2 6 4 = + + 0 10 25 12 21 15 8 12 2 5 6 2 5 9 6 30 3 4 = 0 12 8 10 21 12 25 15 2 1 8 37 = 4 1 42 1 1 8 37 ∴ X = 4 4 1 42 1 37 4 2 4 X = 1 1 21 4 2 2 x 1 1 1 6 4 5 Ex. 4) If + = , find x and y. 3 4 y 3 0 6 12 2 x 1 1 1 6 4 5 Solution: Given + = 3 4 y 3 0 6 12 ∴ 2 x 5 4 5 6 4 y 6 12 ∴ Using definition of equality of matrices, we have 2x = 4, 4y = 12 ∴ x = 2, y = 3 2 a 3 Ex. 5) Find a, b, c if the matrix A = 7 4 5 is a symmetric matrix. c b 6 2 a 3 Solution: Given that A = 7 4 5 is a symmetric matrix. c b 6 ∴ aij = aji for all i and j ∴ a = −7, b = 5, c = 3 #AkashAgrawalTheMathsTutor 44 1 5 Ex. 6) If A = 2 0 Find (AT)T. 3 4 32 1 5 Solution: Let A = 2 0 3 4 32 1 2 3 ∴ AT = 5 0 4 23 1 5 Now (AT)T = 2 0 3 4 32 =A 2 1 2 1 3 1 Ex. 7) If X + Y = 1 3 and X − 2Y = then find X, Y. 3 2 4 2 2 1 2 1 3 1 Solution: Let A = 1 3 and B = 3 2 4 2 X + Y = A......... (1), X − 2Y = B......... (2), Solving (1) and (2) for X and Y Consider (1) − (2), 3Y = A − B, 1 ∴ Y= (A − B) 3 2 1 2 1 1 ∴ Y = 1 3 3 1 3 3 2 4 2 4 2 1 = 2 4 3 7 0 4 2 3 3 = 2 4 3 3 7 0 3 #AkashAgrawalTheMathsTutor 45 3 1 5 a (7) Find a, b, c if b 5 7 is a symmetric matrix. 4 c 0 0 5i x (8) Find x, y, z if y 0 z is a skew symmetric matrix. 3 2 0 2 (9) For each of the following matrices, find its transpose and state whether it is symmetric, skew- symmetric or neither. 1 2 5 2 5 1 0 1 2i i 2 (i) 2 3 4 (ii) 5 4 6 (iii) 1 2i 0 7 5 4 9 1 6 3 2 i 7 0 (10) Construct the matrix A = [aij]3×3 where aij = i − j. State whether A is symmetric or skew symmetric. 1 1 0 1 (11) Solve the following equations for X and Y, if 3X − Y = and X − 3Y = 0 1 1 1 6 6 0 3 2 8 (12) Find matrices A and B, if 2A − B = and A − 2B = 4 2 1 2 1 7 2 x y 1 1 1 6 4 3 5 5 (13) Find x and y, if 3 4 y 4 3 0 3 6 18 7 2a b 3a b 2 3 (14) If , find a, b, c and d. c 2d 2c d 4 1 (15) There are two book shops own by Suresh and Ganesh. Their sales ( in Rupees) for books in three subject - Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B. July sales ( in Rupees), Physics Chemistry Mathematics 5600 6750 8500 A= 6650 First Row Suresh / Second Row Ganesh 7055 8905 August sales ( in Rupees), Physics Chemistry Mathematics 6650 7055 8905 B= 7000 10200 First Row Suresh / Second Row Ganesh 7500 then, (i) Find the increase in sales in Rupees from July to August 2017. (ii) If both book shops get 10% profit in the month of August 2017, find the profit for each book seller in each subject in that month. #AkashAgrawalTheMathsTutor 47 1 3 3 Ex. 2: If A = 3 1 3 find A2 − 5A. What is your conclusion? 3 3 1 Solution : Let A2 = A.A 1 3 3 1 3 3 = 3 1 3 3 1 3 3 3 1 3 3 1 1 9 9 3 3 9 3 9 3 = 3 3 9 9 1 9 9 3 3 3 9 3 9 3 3 9 9 1 19 15 15 = 15 19 15 15 15 19 19 15 15 1 3 3 3 1 3 ∴ A2 − 5A = 15 19 15 − 5 15 15 19 3 3 1 19 15 15 5 15 15 15 19 15 = − 15 5 15 15 15 19 15 15 5 14 0 0 1 0 0 ∴ A2 − 5A = 0 14 0 = 14 0 1 0 = 14 I 0 0 14 0 0 1 ∴ By definition of scalar matrix, A2 − 5A is a scalar matrix. 3 2 Ex. 3: If A = , find k, so that A2 − kA + 2I = O, where I is a 2×2 the identify matrix and 4 2 O is null matrix of order 2. Solution: Given A2 − kA + 2I = O ∴ Here, A2 = AA 3 2 3 2 = 4 2 4 2 9 8 6 4 = 12 8 8 4 1 2 = 4 4 #AkashAgrawalTheMathsTutor 53 ∴ A2 − kA + 2I = O 1 2 3 2 1 0 ∴ − k 4 2 + 2 0 1 = O 4 4 1 2 3k 2k 2 0 0 0 ∴ − + = 4 4 4k 2k 0 2 0 0 1 3k 2 2 2k 0 0 ∴ = 4 4k 4 2k 2 0 0 ∴ Using definition of equality of matrices, we have 1 3k 2 0 3k 3 2 2k 0 2k 2 k=1 4 4k 0 4k 4 4 2k 2 0 2k 2 6 3 4 1 Ex. 4: Find x and y, if 2 0 3 3 1 2 2 1 0 x y 5 4 3 4 6 3 4 1 Solution: Given 2 0 3 3 1 2 2 1 0 x y 5 4 3 4 18 9 8 2 ∴ 2 0 3 3 6 2 0 x y 15 12 6 8 10 7 ∴ 2 0 3 1 6 x y 9 4 ∴ 20 27 14 12 x y ∴ 47 26 x y ∴ x = 47, y = 26 by definition of equality of matrices. (A + B) (A − B) = A2 − AB + BA − B2 Let's Note : Using the distributive laws discussed earlier, we can derive the following results. If A and B are square matrices of the same order, then i) (A + B)2 = A2 + AB + BA + B2 ii) (A − B)2 = A2 − AB − BA + B2 #AkashAgrawalTheMathsTutor 54 1 2 4 3 2 6) If A = , B = 1 0 show that matrix AB is non singular. 1 2 0 1 2 1 2 0 7) If A + I = 5 4 2 , find the product (A + I)(A − I). 0 7 3 1 2 2 8) If A = 2 1 2 , show that A2 – 4A is a scalar matrix. 2 2 1 1 0 9) If A = , find k so that A2 – 8A – kI = O, where I is a 2×2 unit and O is null matrix of 1 7 order 2. 3 1 10) If A = , prove that A2 – 5A + 7I = 0, where I is 2×2 unit matrix. 1 2 1 2 2 a 11) If A = ,B= 1 b and if (A + B)2 = A2 + B2, find value of a and b. 1 2 3 2 12) Find k, If A = and A2 = kA – 2I. 4 2 2 2 1 3 3 3 4 x 13) Find x and y, If 4 1 1 0 2 2 1 1 1 y 2 0 1 1 x 3 1 14) Find x, y, z if 3 0 2 4 1 2 y 1 3 1 2 z 2 2 2 15) Jay and Ram are two friends. Jay wants to buy 4 pens and 8 notebooks, Ram wants to buy 5 pens and 12 notebooks. The price of One pen and one notebook was Rs. 6 and Rs.10 respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks. Properties of the transpose of a matrix: (i) If A and B are two matrices of same order, then (A + B)T = AT + BT (ii) If A is a matrix and k is a constant, then (kA)T = kAT (iii) If A and B are conformable for the product AB, then (AB)T = BT AT #AkashAgrawalTheMathsTutor 56 EXERCISE 2.4 1 3 2 6 1 (1) Find AT, if (i) A = (ii) A = 4 5 4 0 5 (2) If A = [aij]3×3 where aij = 2(i – j). Find A and AT. State whether A and AT both are symmetric or skew symmetric matrices ? 5 3 (3) If A = 4 3 , Prove that (AT)T = A. 2 1 1 2 5 (4) If A = 2 3 4 , Prove that AT = A. 5 4 9 2 3 2 1 1 2 4 1 1 4 (5) If A = 5 4 , B = ,C= then show that 6 1 3 3 2 3 (i) (A + B)T = AT + BT (ii) (A − C)T = AT − CT 5 4 1 3 (6) If A = and B = , then find CT, such that 3A – 2B + C = I, where I is the unit 2 3 4 1 matrix of order 2. 7 3 0 0 2 3 (7) If A = ,B= then find 0 4 2 2 1 4 (i) AT + 4BT (ii) 5AT − 5BT 1 0 1 2 1 4 0 2 3 (8) If A = ,B= and C = 1 1 0 , verify that 3 1 2 3 5 2 (A + 2B + 3C)T = AT + 2BT + 3CT 2 1 1 2 1 (9) If A = and B = 3 2 , prove that (A + BT)T = AT + B. 3 2 3 1 3 (10) Prove that A + A is a symmetric and A − AT is a skew symmetric matrix, where T 1 2 4 5 2 4 3 2 1 (i) A = (ii) A = 3 7 2 2 3 2 4 5 3 (11) Express each of the following matrix as the sum of a symmetric and a skew symmetric matrix. 3 3 1 4 2 (i) (ii) 2 2 1 3 5 4 5 2 #AkashAgrawalTheMathsTutor 59 2 0 1 3) Find the inverse of A = 5 1 0 by using elementary row transformation. 0 1 3 2 0 1 Solution: Let A = 5 1 0 0 1 3 2 0 -1 |A| = 5 1 0 0 1 3 = 2(3 − 0) − 0(15 − 0) − 1(5 − 0) = 6−0−5 = 1≠0 ∴ A−1 is exist. Consider AA−1 = I 2 0 1 1 0 0 5 1 0 −1 0 1 0 A = 0 1 3 0 0 1 6 0 3 3 0 0 5 1 0 A−1 = 0 1 0 By R1 → 3R1 0 1 3 0 0 1 1 1 3 3 1 0 5 1 0 A−1 = 0 1 0 By R1 → R1 − R2 0 1 3 0 0 1 1 1 3 3 1 0 15 6 0 By R2 → R2 − 5R1 0 6 15 A−1 = 0 1 3 0 0 1 1 1 3 3 1 0 By R2 ↔ R3 0 1 3 A−1 = 0 0 1 0 6 16 15 6 0 1 0 0 3 1 1 0 0 1 By R1 → R1 + R2 and R3 → R3 − 6R2 0 1 3 A−1 = 0 0 3 15 6 6 #AkashAgrawalTheMathsTutor 67 M12 = 4, ∴ A12 = (−1)1+2M12 = −1(4) = −4 M21 = −2, ∴ A21 = (−1)2+1M21 = −1(−2) = 2 M22 = 2, ∴ A22 = (−1)2+2M22 = 1(2) = 2 3 4 ∴ Cofactor matrix [Aij]2×2 = 2 2 3 2 adj (A) = [Aij]T = 4 2 1 A−1 = adj (A) A 1 3 2 A−1 = 14 4 2 2 1 1 4) If A = 1 2 1 then find A−1 by the adjoint method. 1 1 2 2 1 1 Solution: Given A = 1 2 1 1 1 2 2 -1 1 |A| = - 1 2 -1 = 2(4 − 1) + 1(−2 + 1) + 1(1 − 2) 1 -1 2 = 6−1−1=4≠0 ∴ A−1 exists. For the given matrix A 2 -1 ∴ A11 = (−1)1+1 = 1(4 − 1) = 3 -1 2 -1 -1 ∴ A12 = (−1)1+2 = −1(−2 + 1) = 1 1 2 -1 2 ∴ A13 = (−1)1+3 = 1(1 − 2) = −1 1 -1 -1 1 ∴ A21 = (−1)2+1 = −1(−2 + 1) = 1 -1 2 #AkashAgrawalTheMathsTutor 70 3) Find the cofactor of the following matrices 5 8 7 1 2 i) ii) 1 2 1 5 8 2 1 1 4) Find the adjoint of the following matrices 1 1 2 2 3 i) ii) 2 3 5 3 5 2 0 1 5) Find the inverse of the following matrices by the adjoint method 1 2 3 3 1 2 2 i) ii) iii) 0 2 4 2 1 4 5 0 0 5 6) Find the inverse of the following matrices by the transformation method. 2 0 1 1 2 i) ii) 5 1 0 2 1 0 1 3 1 0 1 7) Find the inverse A = 0 2 3 by elementary column transformation. 1 2 1 1 2 3 8) Find the inverse 1 1 5 of by the elementary row transformation. 2 4 7 1 0 1 1 2 3 1 1 5 then find matrix X such that XA = B 9) If A = 0 2 3 and B = 1 2 1 2 4 7 1 2 3 1 2 10) Find matrix X, If AX = B where A = 1 1 2 and B = 1 2 4 3 2.7 Applications of Matrices: To find a solution of simultaneous linear equations. Consider the following pair of simultaneous linear equations in two variables. a1 x b1 y c1 .......... (i) a2 x b2 y c2 #AkashAgrawalTheMathsTutor 72 2) Express the following equations in matrix form and solve them by the method of reduction x – y + z = 1, 2x – y = 1, 3x + 3y – 4z = 2. Solution: The given equations can be write as x–y+z=1 2x – y = 1 3x + 3y – 4z = 2 Hence the matrix equation is AX = B 1 1 1 x 1 y ∴ 2 1 0 = 1 3 3 4 z 2 1 1 1 x 1 0 1 2 y 1 By R2 → R2 – 2R1 = 3 3 4 z 2 1 1 1 x 1 0 1 2 y 1 By R3→ R3 – 3R1 = 0 6 7 z 1 1 1 1 x 1 0 1 2 By R3→ R3 – 6R2 y = 1 0 0 5 z 5 We write equations as x – y + z = 1 --------------------------------- (1) y – 2z = −1 ----------------------------------- (2) 5z = 5 ----------------------------------------- (3) From (3), z = 1 Put z = 1 in equation (2) y – 2(1) = −1 ∴y=2–1=1 Put y = 1, z = 1 in equation (1) x – 1 + 1 = 1, ∴x=1 ∴ x = 1, y = 1, z = 1 EXERCISE 2.6 1) Solve the following equations by method of inversion. i) x + 2y = 2, 2x + 3y = 3 ii) 2x + y = 5, 3x + 5y = −3 iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1 iv) x + y + z = 1, x – y + z = 2 and x + y – z = 3 #AkashAgrawalTheMathsTutor 79 2) Express the following equations in matrix form and solve them by method of reduction. i) x + 3y = 2, 3x + 5y = 4 ii) 3x – y = 1 , 4x + y = 6 iii) x + 2y + z = 8, 2x + 3y – z = 11 and 3x – y – 2z = 5 iv) x + y + z = 1, 2x + 3y + 2z = 2 and x + y + 2z = 4 3) The total cost of 3 T.V. and 2 V.C.R. is Rs. 35000. The shopkeeper wants profit of Rs. 1000 per T.V. and Rs. 500 per V.C.R. He sell 2 T.V. and 1 V.C.R. and he gets total revenue as Rs. 21500. Find the cost and selling price of T.V and V.C.R. 4) The sum of the cost of one Economic book, one Co-operation book and one account book is Rs. 420. The total cost of an Economic book, 2 Co-operation books and an Account book is Rs. 480. Also the total cost of an Economic book, 3 Co-operation book and 2 Account books is Rs. 600. Find the cost of each book. Let's Remember Scalar Multiplication of a matrix: If A = [aij]m×n is a matrix and k is a scalar, then kA = [kaij]. Addition of two matrices: Matrices A = [aij] and B = [bij] are said to conformable for addition if orders of A and B are same. Then A + B = [aij + bij]. The order of A + B is the same as the order of A and B. Multiplication of two matrices: A and B are said to be conformable for multiplication if the number of columns of A is equal to the number of rows of B. That is, if A = [aik]m×p and B = [bkj]p×n, then AB is defined and AB = [cij]m×n, p where cij = a k 1 ik bkj i = 1, 2,....., m j = 1, 2,....., n. If A = [aij]m×n is any matrix, then the transpose of A is denoted by AT = B = [bij]n×m and bij = aji If A is a square matrix, then i) A + AT is a symmetric matrix ii) A − AT is a skew-symmetric matrix. Every square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix as 1 1 A = [A + AT] + [A − AT] 2 2 Elementary Transformations: a) Interchange of any two rows or any two columns b) Multiplication of the elements of any row or column by a non zero scalar c) Adding the scalar multiples of all the elements of any row (column) to the corresponding elements of any other row (column). #AkashAgrawalTheMathsTutor 80 IV. Solve the following. 7 3 1) Find k, if is singular matrix. 5 k 2 x 5 2) Find x,y,z if 3 1 z is symmetric matrix. y 5 8 1 5 2 4 2 3 1 5 1 5 3) If A = 7 8 , B = C= then show that (A+B) + C = A + (B+C) 9 5 8 6 7 8 2 5 1 7 4) If A = ,B= 3 0 Find matrix A – 4B + 7I where I is the unit matrix of order 2. 3 7 2 3 3 4 1 5) If A = 3 2 , B = Verify 2 1 3 1 4 i) (A + 2BT)T = AT + 2B ii) (3A − 5BT)T = 3AT − 5B 1 2 3 1 1 1 3 2 1 6) If A = 2 4 6 , B = then show that AB and BA are both singular matrices. 1 2 3 2 1 0 3 1 1 2 7) If A = , B = 5 2 , verify |AB| = |A| |B| 1 5 2 1 8) If A = then show that A2 – 4A + 3I = 0 1 2 3 2 1 a 9) If A = ,B= b 0 and ( A + B )( A – B ) = A2 – B2, find a and b. 2 4 1 2 10) If A = , then find A3 1 3 0 1 2 1 x 1 2 11) Find x, y, z if 5 1 0 3 2 y 1 1 1 1 3 1 2z 2 4 1 1 2 12) If A = 3 2 , B = 2 1 0 then show that (AB) = B A T T T 0 1 #AkashAgrawalTheMathsTutor 84 1 0 0 13) If A = 2 1 0 then reduce it to unit matrix by row transformation. 3 3 1 14) Two farmers Shantaram and Kantaram cultivate three crops rice, wheat and groundnut. The sale (in Rupees.) of these crops by both the farmers for the month of April and May 2016 is given below, April 2016 (In Rs.) Rice Wheat Groundnut Shantaram 15000 13000 12000 Kantaram 18000 15000 8000 May 2016 (In Rs.) Rice Wheat Groundnut Shantaram 18000 15000 12000 Kantaram 21000 16500 16000 Find i) The total sale in rupees for two months of each farmer for each crop. ii)the increase in sale from April to May for every crop of each farmer. 15) Check whether following matrices are invertible or not. 3 4 3 1 2 3 1 0 1 1 i) ii) iii) 1 1 0 iv) 2 4 5 0 1 1 1 2 4 6 1 4 5 16) Find inverse of the following matrices (if they exist) by elementary transformation. 2 3 3 2 0 1 1 1 2 1 i) ii) iii) 2 2 3 iv) 5 1 0 2 3 7 4 3 2 2 0 1 3 3 1 5 17) Find the inverse of 2 7 8 by adjoint method. 1 2 5 18) Solve the following equations by method of inversion. i) 4x – 3y – 2 = 0 , 3x – 4y + 6 = 0 ii) x + y – z = 2 , x – 2y + z = 3 and 2x – y – 3z = −1 iii) x – y + z = 4 , 2x + y – 3z = 0 and x + y + z = 2 19) Solve the following equation by method of reduction. i) 2x + y = 5 , 3x + 5y = −3 ii) x + 2y – z = 3 , 3x – y + 2z = 1 and 2x – 3y + 3z = 2 iii) x – 3y + z = 2 , 3x + y + z = 1 and 5x + y + 3z = 3 #AkashAgrawalTheMathsTutor 85 3 Differentiation dy du dv Let's Study 2. y = u − v then = − dx dx dx 1. Derivatives of composite functions. dy dv du 3. y = u.v then =u + v 2. Derivatives of inverse functions. dx dx dx 3. Derivatives of logarithmic functions. du dv u dy v −u 4. y= then = dx dx v ≠ 0 4. Derivatives of implicit function. v dx 2 v 5. Derivatives of parametric functions. dy du 6. Derivative of second order. 5. y = k. u then = k. , k constant. dx dx Let's Recall Introduction: 1. Concept of continuity In Standard XI, we have studied the concept of differentiation. We have used this concept in 2. Concept of Differentiability. calculating marginal demand and marginal cost 3. Derivatives of some standard functions. of a commodity. dy y = f(x) = f '( x) dx Let's Learn 1 K(constant) 0 2 x 1 3.1 Derivative of a Composite Function: 3 Sometimes complex looking functions x 1 can be greatly simplified by expressing 2 x them as compositions of two or more 4 1 −1 different functions. It is then not possible x x2 to differentiate them directly is possible with simple functions. 5 xn n.xn−1 6 ax ax.loga Now, we discuss differentiation of such composite functions using the chain rule. 7 ex ex 8 logx Result 1: If y = f(u) is a differentiable function 1 of u and u = g(x) is a differentiable function x of x then 4. Rules of Differentiation: dy = dy × du If u and v are differentiable functions of x dx du dx and if (This is called Chain Rule) dy du dv 1. y = u + v then = + dx dx dx #AkashAgrawalTheMathsTutor 89 EXERCISE 3.1 SOLVED EXAMPLES 1) Find rate of change of demand (x) of a dy Q.1 Find if, commodity with respect to its price (y) if dx y = 20 + 15x + x2 1 1) y= x+ Solution: Let y = 20 + 15x + x2 x Differentiating both sides with respect to x, 2) y= 3 a +x2 2 we get 3) y = (5x3 − 4x2 − 8x)9 dy ∴ = 15 + 2x dx dy By derivative of the inverse function Q.2 Find if, dx dx 1 dy ∴ = dy , ≠0 1) y = log(logx) dy dx dx 2) y = log(10x4 + 5x3 − 3x2 + 2) ∴ rate of change of demand with respect 3) y = log(ax2 + bx + c) dx 1 dy to price = = Q.3 Find if, dy 15 + 2 x dx 5x 2 x 4 2) Find rate of change of demand (x) of a y= e 2 1) commodity with respect to its price (y) if (1+ log x ) 2) y= a y = 5 + x2e −x + 2x ( x + log x ) 3) y= 5 Solution: Let y = 5 + x2e −x + 2x Differentiating both sides with respect to x, 3.2 Derivative of an Inverse Function: we get Let y = f(x) be a real valued function dy defined on an appropriate domain. The ∴ = (−x2e−x + 2xe−x + 2) inverse of this function exists if and only if dx the function is one-one and onto. By derivative of the inverse function dx 1 dy For example: Let f : R → R be such that ∴ = dy , ≠0 dy dx f(x) = x +10 then inverse of f is dx f −1 : R → R such that f −1(y) = y −10 ∴ Rate of change of demand with respect That is, if y = x +10 then x = y − 10 dx 1 to price = = ( x 2 e x 2 xe x 2) Result 2 : If y = f(x) is a differentiable function dy of x such that inverse function x = f−1(y) exists, then x is a differentiable function of 3) Find rate of change of demand (x) of a y commodity with respect to its price (y) if dx 1 3x + 7 dy y= and = dy , ≠0 2x2 + 5 dy dx dx
