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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise

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 [โ– 8(1&3@2&6)] [โ– 8(๐‘ฅ@๐‘ฆ)] = [โ– 8(5@8)] Hence A = [โ– 8(1&3@2&6)] , X = [โ– 8(๐‘ฅ@๐‘ฆ)] & B = [โ– 8(5@8)] Calculating |A| |A| = |โ– 8(1&3@2&6)| = 1(6) โ€“ 2(3) = 6 โ€“ 6 = 0 Since |A| = 0, we will calculate (adj A) B A = [โ– 8(1&3@2&6)] adj (A) = [โ– 8(1&3@2&6)] = [โ– 8(6&โˆ’3@โˆ’2&1)] Now, (adj A)(B) = [โ– 8(6&โˆ’3@โˆ’2&1)] [โ– 8(5@8)] = [โ– 8(6(5)+(โคถ7โˆ’3)8@โˆ’2(5)+1(8))] = [โ– 8(30โˆ’24@โˆ’10+8)] = [โ– 8(6@โˆ’2)] โ‰  O Since adj A (B) โ‰  O & |A| = 0, , There exists no solution, Hence, the given system of equation is inconsistent

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.