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Ex 4.6, 5 - Examine consistency - Chapter 4 Determinants

Ex 4.6, 5 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.6, 5 - Chapter 4 Class 12 Determinants - Part 3 Ex 4.6, 5 - Chapter 4 Class 12 Determinants - Part 4 Ex 4.6, 5 - Chapter 4 Class 12 Determinants - Part 5 Ex 4.6, 5 - Chapter 4 Class 12 Determinants - Part 6

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Ex 4.6, 5 Examine the consistency of the system of equations. 3x − y − 2z = 2 2y − z = −1 3x − 5y = 3 The system of equations can be written as 3x − y − 2z = 2 0x + 2y − z = −1 3x − 5y + 0z = 3 Writing equation as AX = B [■8(3&−1&−[email protected]&2&−[email protected]&−5&0)] [■8(𝑥@𝑦@𝑧)] = [■8([email protected]@3)] Hence A = [■8(3&−1&−[email protected]&2&−[email protected]&−5&0)], X = [■8(𝑥@𝑦@𝑧)] & B = [■8([email protected]@3)] Calculating |A| |A| = |■8(3&−1&−[email protected]&2&−[email protected]&−5&0)| = 3 |■8(2&−[email protected]−5&0)| – (–1) |■8(0&−[email protected]&0)| − 2 |■8(0&[email protected]&−5)| = 3(0 – 5) + 1 (0 + 3) –2 (0 – 6)= 3 (–5) + 1 (3) –2 (–6) = –15 + 3 + 12 = –15 + 15 = 0 Since |A| = 0, We calculate adj A (B) adj A = [■8(A_11&A_12&[email protected]_21&A_22&[email protected]_31&A_32&A_33 )]^′= [■8(A_11&A_21&[email protected]_12&A_22&[email protected]_13&A_32&A_33 )] A = [■8(3&−1&−[email protected]&2&−[email protected]&−5&0)] M11 = |■8(2&−[email protected]−5&0)| = 0 – (5) = – 5 M12 = |■8(0&−[email protected]&0)| = 0 + 3 = –3 M13 = |■8(6&[email protected]&−5)| = 0 – 6 = – 6 M21 = |■8(−1&−[email protected]−5&0)| = 0 – 10 = – 10 M22 = |■8(3&−[email protected]&0)| = 0 + 6 = 6 M23 = |■8(3&−[email protected]&−5)| = – 15 +3 = – 12 M31 = |■8(−1&−[email protected]&−1)| = 1 + 4 = 5 M32 = |■8(3&−[email protected]&−1)| = – 3 + 0 = – 3 M33 = |■8(3&−[email protected]&2)| = 6 + 0 = 6 Now, A11 = (–1)1+1 . M11 = (–1)2 . (–5) = –5 A12 = (–1)1+2 . M12 = (–1)3 . 3 = –3 A13 = (–1)1+3 . M13 = (–1)4 . (–6) = –6 A21 = (–1)2+1 . M21 = (–1)3 . (–10) = 10 A22 = (–1)2+2 . M22 = (–1)4 . 6 = 6 A23 = (–1)2+3 . M23 = (–1)5 . (–12) = 12 A31 = (–1)3+1 . M31 = (–1)4 . 5 = 5 A32 = (–1)3+2 . M32 = (–1)5 . (–3) = 3 A33 = (–1)3+3 . M33 = (–1)6 . (6) = 6 Thus, adj (A) = [■8(A_11&A_21&[email protected]_12&A_22&[email protected]_13&A_32&A_33 )] = [■8(3&−1&−[email protected]&2&−[email protected]&−5&0)] Now, adj (A) . B Putting values = [■8(−5&10&[email protected]−3&6&[email protected]−6&12&6)] [■8([email protected][email protected])] = [■8(−5(2)+10(−1)+5(3)@−3(2)+6(−1)+3(3)@−6(2)+12(−1)+6(3) )] = [■8(−10−[email protected]−6−[email protected]−12−12+18)] = [■8(−[email protected][email protected]−6)] Thus, adj A . B ≠ O Since |A| = 0 & (adj A) B ≠ O, Thus, the given system equation is inconsistent & the system of equations has no solution

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Davneet Singh

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.