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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

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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&[email protected]&3)] [β– 8(π‘₯@𝑦)] = [β– 8([email protected])] Hence A = [β– 8(5&[email protected]&3)], X = [β– 8(π‘₯@𝑦)] & B = [β– 8([email protected])] Calculating |A| |A| = |β– 8(5&[email protected]&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&[email protected]&3)] adj A = [β– 8(5&[email protected]&3)] = [β– 8(3&βˆ’[email protected]βˆ’7&5)] Now, A-1 = 1/(|A|) adj A Putting values = 1/1 [β– 8(3&βˆ’[email protected]βˆ’7&5)] = [β– 8(3&βˆ’[email protected]βˆ’7&5)] Now X = A-1 B [β– 8(π‘₯@𝑦)] = [β– 8(3&βˆ’[email protected]βˆ’7&5)] [β– 8([email protected])] [β– 8(π‘₯@𝑦)] = [β– 8(3(4)+(•7βˆ’2)[email protected]βˆ’7(4)+5(5))] [β– 8(π‘₯@𝑦)] = [β– 8([email protected]βˆ’3)] Hence, x = 2 & y = –3

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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.