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


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Ex 4.4, 9 Find the inverse of each of the matrices (if it exists). [■8(2&1&3@4&−1&0@−7&2&1)] Let A = [■8(2&1&3@4&−1&0@−7&2&1)] We know that A–1 = 1/(|A|) (adj A) exists if |A|≠ 0 Calculating |A| |A| = |■8(2&1&3@4&−1&0@−7&2&1)| = 2 |■8(–1&0@2&1)| – 1 |■8(4&0@−7&1)| + 3 |■8(4&−1@−7&2)| = 2(–1 – 0) – 1 (4 – 0) + 3 (8 – 7) = 2 (–1) – 1 (4) + 3 (1) = –3 Since |𝐴| ≠ 0 , A–1 exists Calculating adj (A) adj (A) = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(2&1&3@4&−1&0@−7&2&1)] M11 = |■8(−1&0@2&1)| = –1(1) – 0(2) = -1 M12 = |■8(4&0@−7&1)| = 4(1) – 0(–7) = 4 M13 = |■8(4&−1@−7&2)| = 4(2) – (-7)(-1) = 1 M21 = |■8(1&3@2&1)| = 1(1) – 2(3) = – 5 M22 = |■8(2&3@−7&1)| = 2(1) – (−7)(3)= 23 M23 = |■8(2&1@−7&2)| = 2(2) – (−7)(1) = 11 M31 = |■8(1&3@−1&0)| = 1(0) – (−1)(3) = 3 M32 = |■8(2&3@4&0)| = 2(0) – 4(3) = –12 M33 = |■8(2&1@4&−1)| = 2(-1) – 4(1) = – 6 Now, A11 = (–1)1 + 1 M11 = (–1)2 (–1) = 1 (–1) = – 1 A12 = (–1)1+2 M12 = (–1)3 ( 4) = ( –1) (4) = –4 A13 = (–1)1+3 M13 = (–1)4 (1) = 1 A21 = (–1)2+1 M21 = (–1)3 ( –5) = (−1)(−5) = 5 A22 = (–1)2+2 M22 = (–1)4 ( –23) = 23 A23 = (–1)2+3 M23 = (–1)5 11 = −1 (11) = – 11 A31 = (–1)3+1 M31 = (–1)4 (4) = 1 (3) = 3 A32 = (–1)3+2 M32 = (–1)5 (–12) = (–1) (–12) = 12 A33 = (–1)3+3 M33 = (–1)6 (–6) = 1 . (–6) = – 6 Thus, adj (A) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] = [■8(−1&5&3@−4&23&12@1&−11&−6)] Calculating inverse A– 1 = 1/(|A|) ( adj (A)) = 1/(−3) [■8(−1&5&3@−4&23&12@1&−11&−6)] = (−𝟏)/𝟑 [■8(−𝟏&𝟓&𝟑@−𝟒&𝟐𝟑&𝟏𝟐@𝟏&−𝟏𝟏&−𝟔)]

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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.