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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.