Check sibling questions

Example 24 - Verify that A adj A = |A| I. Also find A-1 - Examples

Example 24 - Chapter 4 Class 12 Determinants - Part 2
Example 24 - Chapter 4 Class 12 Determinants - Part 3
Example 24 - Chapter 4 Class 12 Determinants - Part 4
Example 24 - Chapter 4 Class 12 Determinants - Part 5
Example 24 - Chapter 4 Class 12 Determinants - Part 6

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Transcript

Example 24 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.