Question 2 - Case Based Questions (MCQ) - Chapter 4 Class 12 Determinants

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Question The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to kept the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. The sum of the number of awardees for honesty and supervision is twice the number of awardees for helping. Given that Awards for honesty = x Awards for helping (cooperation) = y Awards for supervising = z Since the sum of all the awardees is 12 x + y + z = 12 Also, Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33 3(y + z) + 2x = 33 2x + 3y + 3z = 33 And, The sum of the number of awardees for honesty and supervision is twice the number of awardees for helping. x + z = 2y x −2y + z = 0 Question 1 x + y + z = _______. (a) 3 (b) 5 (c) 7 (d) 12 From (1) x + y + z = 12 So, the correct answer is (D) Question 2 x – 2y = _______. (a) z (b) – z (c) 2z (d) –2z From (3) x − 2y + z = 0 x − 2y = −z So, the correct answer is (B) Question 3 The value of z is _______. (a) 3 (b) 4 (c) 5 (d) 6 Now, our equations are x + y + z = 12 2x + 3y + 3z = 33 x − 2y + z = 0 Writing equation as AX = B [■8(1&1&1@2&3&3@1&−2&1)] [■8(𝑥@𝑦@𝑧)] = [■8(12@33@0)] Hence A = [■8(1&1&1@2&3&3@1&−2&1)], X = [■8(𝑥@𝑦@𝑧)] & B = [■8(12@33@0)] Calculating |A| |A| = |■8(1&1&1@2&3&3@1&−2&1)| = 1 |■8(3&3@−2&1)| – 1 |■8(2&3@1&1)| + 1 |■8(2&3@1&−2)| = 1(3 + 6) − 1 (2 − 3) + 1 (−4 − 3) = 1 (9) − 1 (−1) + 1 (–7) = 9 + 1 − 7 = 3 Thus, |A| ≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) adj (A) = [■8(A_11&A_12&A_13@A_21&A_22&A_23@A_31&A_32&A_33 )]^′ = [■8(A_11&A_21&A_31@A_12&A_22&A_23@A_13&A_32&A_33 )] And, A = [■8(1&1&1@2&3&3@1&−2&1)] M11 = [■8(3&3@−2&1)] = 3 + 6 = 9 M12 = [■8(2&3@1&1)] = 2 − 3 = −1 M13 = [■8(2&3@1&−2)] = −4 − 3 = −7 M21 = [■8(1&1@−2&1)] = 1 + 2 = 3 M22 = [■8(1&1@1&1)] = 1 − 1 = 0 M23 = [■8(1&1@1&−2)] = −2 − 1 = −3 M31 = [■8(1&1@3&3)] = 3 − 3 = 0 M32 = [■8(1&1@2&3)] = 3 − 2 = 1 M33 = [■8(1&1@2&3)] = 3 − 2 = 1 Now, A11 = (–1)1+1 . M11 = (–1)2 . (9) = 9 A12 = (–1)1+2 . M12 = (–1)3 . (–1) = 1 A13 = (–1)1+3 . M13 = (–1)4 . (−7) = −7 A21 = (–1)2+1 . M21 = (–1)3 . (3) = −3 A22 = (–1)2+2 . M22 = (–1)4 . (0) = 0 A23 = (–1)2+3 . M23 = (–1)5 . (−3) = 3 A31 = (–1)3+1 . M31 = (–1)4 . (0) = 0 A32 = (–1)3+2 . M32 = (–1)5 . (1) = −1 A33 = (–1)3+3 . M33 = (–1)6 . (1) = 1 Thus, adj (A) =[■8(9&−3&0@1&0&−1@−7&3&1)] Now, A-1 = 1/(|A|) adj A Putting values = 1/3 [■8(9&−3&0@1&0&−1@−7&3&1)] Also, X = A-1 B Putting values [█(■8(𝑥@𝑦)@𝑧)] = 1/3 [■8(9&−3&0@1&0&−1@−7&3&1)][■8(12@33@0)] [█(■8(𝑥@𝑦)@𝑧)] = 1/3 [■8(9(12)−3(33)+0(0)@1(12)+0(33)+(−1)(0)@−7(12)+3(33)+1(0) )] [█(■8(𝑥@𝑦)@𝑧)] = 1/3 [■8(108−99@12@−84+99)] = 1/3 [█(■8(9@12)@15)] [█(■8(𝑥@𝑦)@𝑧)] = [█(■8(3@4)@5)] Hence, x = 3, y = 4 & z = 5 Since z = 5 So, the correct answer is (C) Question 4 The value of x + 2y = _______. (a) 9 (b) 10 (c) 11 (d) 12 Since, x = 3, y = 4 & z = 5 Thus, x + 2y = 3 + 2(4) = 3 + 8 = 11 So, the correct answer is (C) Question 5 The value of 2x + 3y + 5z = _______. (a) 40 (b) 43 (c) 50 (d) 53 Since, x = 3, y = 4 & z = 5 Thus, 2x + 3y + 5z = 2(3) + 3(4) + 5(5) = 6 + 12 + 25 = 43 So, the correct answer is (B)