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Example 13 If A = [■8(1&3&3@1&4&3@1&3&4)], then verify that A adj A = |A| I. Also find A–1. Solving L.H.S A (adj A) First Calculating adj A adj A = [■8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_23&A_33 )] Now, A = [■8(1&3&3@1&4&3@1&3&4)] M11 = |■8(4&3@3&4)| = 4(4) – 3(3) = 7 M12 = |■8(1&3@1&4)| = 1(4) – 1(3) = 1 M13 = |■8(1&4@1&3)| = 1(3) – 1(4) = –1 M21 = |■8(3&3@3&4)| = 3(4) – 3(3) = 3 M22 = |■8(1&3@1&4)| = 1(4) – 1(3) = 1 M23 = |■8(1&3@1&3)| = 1(3) – 1(3) = 0 M31 = |■8(3&3@4&3)| = 3(3) – 4(3) = – 3 M32 = |■8(1&3@1&3)| = 1(3) – 1(3) = 0 M33 = |■8(1&3@1&4)| = 1(4) – 1(3) = 1 Now, A11 = (–1)1 + 1 M11 = (–1)2 7 = 7 A12 = (–1)1+2 M12 = (–1)3 (1) = –1 A13 = (–1)1+3 M13 = (–1)4 (–1) = –1 A21 = (–1)2+1 M21 = (–1)3 (3) = –3 A22 = (–1)2+2 M22 = (–1)4 (1) = 1 A23 = (–1)2+3 M23 = (–1)5 0 = 0 A31 = (–1)3+1 M31 = (–1)4 (– 3) = –3 A32 = (–1)3+2 M32 = (–1)5 0 = 0 A33 = (–1)3+3 M33 = (–1)6 (1) = 1 Thus, adj (A) = [■8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_23&A_33 )] = [■8(7&−3&−3@−1&1&0@−1&0&1)] Finding A (adj A) A adj (A) = [■8(1&3&3@1&4&3@1&3&4)] [■8(7&3&−3@−1&1&0@−1&0&1)] = [■8(1(7)+3(⤶7−1)+3(−1)&1(−3)+3(1)+3(0)&1(−3)+3(0)+3(1)@1(7)+4(⤶7−1)+3(−1)&1(−3)+4(1)+3(0)&1(−3)+4(0)+3(1)@1(7)+3(⤶7−1)+4(−1)&1(−3)+3(1)+4(0)&1(−3)+3(0)+4(1))] = [■8(7−3−3&−3+3+0&−3+0+3@7−4−3&−3+4+0&−3+0+3@7−3−4&−3+3+0&−3+0+4)] = [■8(𝟏&𝟎&𝟎@𝟎&𝟏&𝟎@𝟎&𝟎&𝟏)] Solving R.H.S |A| I Calculating |A| |A| = |■8(1&3&3@1&4&3@1&3&4)| = 1 (4(4) – 3(3)) – 3(1(4) – 1(3)) + 3(1(4) – 1(3)) = 1(7) – 3(1) +3( – 1) = 7 – 3 – 3 = 1 Now, |A| I = 1 [■8(1&0&0@0&1&0@0&0&1)] = [■8(1&0&0@0&1&0@0&0&1)] = L.H.S Thus, A(adj A) = |A| I Hence proved Finding A-1 We know that A-1 = 1/(|A|) (adj A) exists if |A| ≠ 0 Here, |A| = 1 ≠ 0 Thus A-1 exists So, A-1 = 1/(|A|) (adj A) = 1/1 [■8(7&−3&−3@−1&1&0@−1&0&1)] = [■8(𝟕&−𝟑&−𝟑@−𝟏&𝟏&𝟎@−𝟏&𝟎&𝟏)]

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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