Examples

Chapter 4 Class 12 Determinants
Serial order wise

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Example 24 If A = [■8(1&3&[email protected]&4&[email protected]&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&[email protected]_12&A_22&[email protected]_13&A_23&A_33 )] Now, A = [■8(1&3&[email protected]&4&[email protected]&3&4)] M11 = |■8(4&[email protected]&4)| = 4(4) – 3(3) = 7 M12 = |■8(1&[email protected]&4)| = 1(4) – 1(3) = 1 M13 = |■8(1&[email protected]&3)| = 1(3) – 1(4) = –1 M21 = |■8(3&[email protected]&4)| = 3(4) – 3(3) = 3 M22 = |■8(1&[email protected]&4)| = 1(4) – 1(3) = 1 M23 = |■8(1&[email protected]&3)| = 1(3) – 1(3) = 0 M31 = |■8(3&[email protected]&3)| = 3(3) – 4(3) = – 3 M32 = |■8(1&[email protected]&3)| = 1(3) – 1(3) = 0 M33 = |■8(1&[email protected]&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&[email protected]_12&A_22&[email protected]_13&A_23&A_33 )] = [■8(7&−3&−[email protected]−1&1&[email protected]−1&0&1)] Finding A (adj A) A adj (A) = [■8(1&3&[email protected]&4&[email protected]&3&4)] [■8(7&3&−[email protected]−1&1&[email protected]−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&−[email protected]−4−3&−3+4+0&−[email protected]−3−4&−3+3+0&−3+0+4)] = [■8(1&0&[email protected]&1&[email protected]&0&1)] Solving R.H.S |A| I Calculating |A| |A| = |■8(1&3&[email protected]&4&[email protected]&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&[email protected]&1&[email protected]&0&1)] = [■8(1&0&[email protected]&1&[email protected]&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&−[email protected]−1&1&[email protected]−1&0&1)] = [■8(𝟕&−𝟑&−𝟑@−𝟏&𝟏&𝟎@−𝟏&𝟎&𝟏)]

#### 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.