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Ex 4.6, 3 - Examine consistency x + 3y = 5, 2x + 6y = 8 - Ex 4.6

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


Transcript

Ex 4.6, 3 Examine the consistency of the system of equations. x + 3y = 5 2x + 6y = 8 x + 3y = 5 2x + 6y = 8 Writing equation as AX = B [β– 8(1&3@2&6)] [β– 8(π‘₯@𝑦)] = [β– 8(5@8)] Hence A = [β– 8(1&3@2&6)] , X = [β– 8(π‘₯@𝑦)] & B = [β– 8(5@8)] Calculating |A| |A| = |β– 8(1&3@2&6)| = 1(6) – 2(3) = 6 – 6 = 0 Since |A| = 0, we will calculate (adj A) B A = [β– 8(1&3@2&6)] adj (A) = [β– 8(1&3@2&6)] = [β– 8(6&βˆ’3@βˆ’2&1)] Now, (adj A)(B) = [β– 8(6&βˆ’3@βˆ’2&1)] [β– 8(5@8)] = [β– 8(6(5)+(•7βˆ’3)8@βˆ’2(5)+1(8))] = [β– 8(30βˆ’24@βˆ’10+8)] = [β– 8(6@βˆ’2)] β‰  O Since adj A (B) β‰  O & |A| = 0, , There exists no solution, Hence, the given system of equation is inconsistent

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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.