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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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.