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Ex 4.4, 8 Find the inverse of each of the matrices (if it exists). [■8(1&0&0@3&3&0@5&2&−1)] Let A = [■8(1&0&0@3&3&0@5&2&−1)] We know that A–1 = 1/(|A|) (adj A) exists if |A|≠ 0 Step 1: Calculate |A| |A| = |■8(1&0&0@3&3&0@5&2&−1)| = 1 |■8(3&0@2&−1)| – 0 |■8(3&0@5&1)| + 0|■8(3&3@5&2)| = 1(–3 – 0) + 0 + 0 = −3 Since |A| ≠ 0 , A–1 exists Step 2: Calculate adj A adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] A = [■8(1&0&0@3&3&0@5&2&−1)] M11 = |■8(3&0@2&−1)| = 3(−1) – 2(0) = −3 M12 = |■8(3&0@5&−1)| = 3(–1) – 0(5) = – 3 M13 = |■8(3&3@5&2)| = 3(2) – 5(3) = – 9 M21 = |■8(0&0@2&–1)| = 0(–1) – 2(0) = 0 M22 = |■8(1&0@5&−1)| = 1(– 1) – 0 = – 1 M23 = |■8(1&0@5&2)| = 2(1) – 5(0) = 2 M31 = |■8(0&0@3&0)| = 0(0) – 3(0) = 0 M32 = |■8(1&0@3&0)| = 1(0) – 3(0) = 0 M33 = |■8(1&0@3&3)| = 1(3) – 3(0) = 3 A11 = ( – 1)1 + 1 M11 = ( – 1)2 (– 3) = 1 ( – 3) = – 3 A12 = ( – 1)1+2 M12 = ( – 1)3 ( – 3) = ( – 1) (– 3) = 3 A13 = ( – 1)1+3 M13 = ( – 1)4 ( – 9) = 1 . (– 9) = – 9 A21 = ( – 1)2+1 M21 = ( – 1)3 (0) = 0 A22 = ( – 1)2+2 M22 = ( – 1)4 ( – 1) = 1 . ( – 1) = −1 A23 = ( – 1)2+3 M23 = ( – 1)5 (2) = – 1 (2) = – 2 A31 = ( – 1)3+1 M31 = ( – 1)4 (2) = 0 A32 = ( – 1)3+2 M32 = ( – 1)5 (0) = 0 A33 = ( – 1)3+3 M33 = ( – 1)6 3 = 3 ∴ adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] = [■8(−3&0&0@3&−1&0@−9&−2&3)] Step 3: Calculate A–1 A– 1 = 1/(|A|) ( adj (A)) = 1/(−3) [■8(−3&0&0@3&−1&0@−9&−2&3)] = (−𝟏)/𝟑 [■8(−𝟑&𝟎&𝟎@𝟑&−𝟏&𝟎@−𝟗&−𝟐&𝟑)]

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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