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Ex 4.6, 10 - Solve using matrix method 5x+2y=3 3x+2y=5 - Ex 4.6

Ex 4.6, 10 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.6, 10 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.6, 10 - Chapter 4 Class 12 Determinants - Part 4

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Ex 4.6, 10 Solve system of linear equations, using matrix method. 5x + 2y = 3 3x + 2y = 5 The system of equation is 5x + 2y = 3 3x + 2y = 5 Writing above equation as AX = B [β– 8(5&2@3&2)] [β– 8(π‘₯@𝑦)] = [β– 8(3@5)] Hence A = [β– 8(5&2@3&2)], X = [β– 8(π‘₯@𝑦)] & B = [β– 8(3@5)] Calculating |A| |A| = |β– 8(5&2@3&2)| = 5(2) –3(2) = 10 – 6 = 4 Since |A| β‰  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) A =[β– 8(5&2@3&2)] adj A = [β– 8(2&βˆ’2@βˆ’3&5)] Now, A-1 = 1/(|A|) adj A A-1 = 1/4 [β– 8(2&βˆ’2@βˆ’3&5)] Thus, X = A-1 B [β– 8(π‘₯@𝑦)] = 1/4 [β– 8(2&βˆ’2@βˆ’3&5)] [β– 8(3@5)] = 1/4 [β– 8(2(3)+(•7βˆ’2)5@βˆ’3(3)+5 (5))] [β– 8(π‘₯@𝑦)] = 1/4 [β– 8(6βˆ’10@βˆ’9+25)] = 1/4 [β– 8(βˆ’4@16)] [β– 8(π‘₯@𝑦)] = [β– 8(βˆ’1@4)] Hence, x = –1 & y = 4

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.