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Ex 4.5, 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&−2@0&2&−1@3&−5&0)] [■8(𝑥@𝑦@𝑧)] = [■8(2@1@3)] Hence A = [■8(3&−1&−2@0&2&−1@3&−5&0)], X = [■8(𝑥@𝑦@𝑧)] & B = [■8(2@1@3)] Calculating |A| |A| = |■8(3&−1&−2@0&2&−1@3&−5&0)| = 3 |■8(2&−1@−5&0)| – (–1) |■8(0&−1@3&0)| − 2 |■8(0&2@3&−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&A_13@A_21&A_22&A_23@A_31&A_32&A_33 )]^′= [■8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_32&A_33 )] A = [■8(3&−1&−2@0&2&−1@3&−5&0)] M11 = |■8(2&−1@−5&0)| = 0 – (5) = – 5 M12 = |■8(0&−1@3&0)| = 0 + 3 = –3 M13 = |■8(6&2@3&−5)| = 0 – 6 = – 6 M21 = |■8(−1&−2@−5&0)| = 0 – 10 = – 10 M22 = |■8(3&−2@3&0)| = 0 + 6 = 6 M23 = |■8(3&−1@3&−5)| = – 15 +3 = – 12 M31 = |■8(−1&−2@2&−1)| = 1 + 4 = 5 M32 = |■8(3&−2@0&−1)| = – 3 + 0 = – 3 M33 = |■8(3&−1@0&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&A_31@A_12&A_22&A_32@A_13&A_32&A_33 )] = [■8(3&−1&−2@0&2&−1@3&−5&0)] Now, adj (A) . B Putting values = [■8(−5&10&5@−3&6&3@−6&12&6)] [■8(2@−1@3)] = [■8(−5(2)+10(−1)+5(3)@−3(2)+6(−1)+3(3)@−6(2)+12(−1)+6(3) )] = [■8(−10−10+15@−6−6+9@−12−12+18)] = [■8(−5@−3@−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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo