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Ex 4.6, 7 - Solve using matrix method 5x + 2y = 4, 7x + 3y = 5

Ex 4.6, 7 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.6, 7 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.6, 7 - Chapter 4 Class 12 Determinants - Part 4


Transcript

Ex 4.6, 7 Solve system of linear equations, using matrix method. 5x+ 2y = 4 7x + 3y = 5 The system of equations is 5x + 2y = 4 7x + 3y = 5 Writing equation as AX = B [β– 8(5&2@7&3)] [β– 8(π‘₯@𝑦)] = [β– 8(4@5)] Hence A = [β– 8(5&2@7&3)], X = [β– 8(π‘₯@𝑦)] & B = [β– 8(4@5)] Calculating |A| |A| = |β– 8(5&2@7&3)| = 5(3) – 7(2) = 15 – 14 = 1 β‰  0 Since |A|β‰  0, System of equations is consistent & has a unique solution Now, AX = B Solution is X = A-1 B Calculating Aβˆ’1 A-1 = 1/(|A|) adj (A) A = [β– 8(5&2@7&3)] adj A = [β– 8(5&2@7&3)] = [β– 8(3&βˆ’2@βˆ’7&5)] Now, A-1 = 1/(|A|) adj A Putting values = 1/1 [β– 8(3&βˆ’2@βˆ’7&5)] = [β– 8(3&βˆ’2@βˆ’7&5)] Now X = A-1 B [β– 8(π‘₯@𝑦)] = [β– 8(3&βˆ’2@βˆ’7&5)] [β– 8(4@5)] [β– 8(π‘₯@𝑦)] = [β– 8(3(4)+(•7βˆ’2)5@βˆ’7(4)+5(5))] [β– 8(π‘₯@𝑦)] = [β– 8(2@βˆ’3)] Hence, x = 2 & y = –3

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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.