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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 4.5, 9 Solve system of linear equations, using matrix method. 4x – 3y = 3 3x – 5y = 7 The system of equation is 4x – 3y = 3 3x – 5y = 7 Writing equation as AX = B [■8(4&−3@3&−5)] [■8(𝑥@𝑦)] = [■8(3@7)] Hence A = [■8(4&−3@3&−5)], X = [■8(𝑥@𝑦)] & B = [■8(3@7)] Calculating |A| |A| = |■8(4&−3@3&−5)| = 4(–5) – 3 (–3) = –20 + 9 = – 11 Since |A| ≠ 0 ∴ 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) A = [■8(4&−3@3&−5)] adj A = [■8(4&−3@3&−5)] = [■8(−5&3@−3&4)] Now, A-1 = 1/(|A|) adj A A-1 = 1/(−11) [■8(−5&3@−3&4)] = 1/11 [■8(5&−3@3&−4)] Thus, X = A-1 B [■8(𝑥@𝑦)] = 1/11 [■8(5&−3@3&−4)] [■8(3@7)] [■8(𝑥@𝑦)] = 1/11 [■8(5(3)+(⤶7−3)7@3(3)+(−4)7)] = 1/11 [■8(15−21@9−28)] [■8(𝑥@𝑦)] = [((−6)/11)¦((−19)/11)] Hence, x = (−𝟔)/𝟏𝟏 & y = (−𝟏𝟗)/𝟏𝟏

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.