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Ex 4.5, 4 - Verify A (adj A) = (adj A) A =  AI - Class 12 CBSE

Ex 4.5, 4 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.5, 4 - Chapter 4 Class 12 Determinants - Part 3 Ex 4.5, 4 - Chapter 4 Class 12 Determinants - Part 4 Ex 4.5, 4 - Chapter 4 Class 12 Determinants - Part 5 Ex 4.5, 4 - Chapter 4 Class 12 Determinants - Part 6

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Ex 4.4, 4 Verify A (adj A) = (adj A) A = |𝐴|I for A = [■8(1&−1&[email protected]&0&−[email protected]&0&3)] Calculating |𝑨| |A| = |■8(1&−1&[email protected]&0&−[email protected]&0&3)| = 1 |■8(0&−[email protected]&3)| – (–1) |■8(3&−[email protected]&3)| +2 |■8(3&[email protected]&0)| = 1 (0 – 0) + 1 (9 + 2) +2 (0 – 0) = 11 Calculating adj A adj A = [■8(A_11&A_21&[email protected]_12&A_22&[email protected]_13&A_23&A_33 )] A = [■8(1&−1&[email protected]&0&−[email protected]&0&3)] M11 = |■8(0&−[email protected]&3)| = 0(3) – 0(–2) = 0 M12 = |■8(3&−[email protected]&3)| = 3(3) – 1(–2) = 11 M13 = |■8(3&[email protected]&0)| = 3(0) – 0(1) = 0 M21 = |■8(−1&[email protected]&3)| = −1(3) – 0(2) = −3 M22 = |■8(1&[email protected]&3)| = 1(3) – 1(2) = 1 M23 = |■8(1&−[email protected]&0)| = 1(0) – 1(−1) = 1 M31 = |■8("−" 1&[email protected]&"−" 2)| = -1(–2) – 0(2) = 2 M32 = |■8(1&[email protected]&−2)| = 1(–2) – 3(2) = –8 M33 = |■8(1&−[email protected]&0)| = 1(0) – 3(−1) = 3 Now, A11 = (–1)1 + 1 M11 = (–1)2 0 = 0 A12 = (–1)1+2 M12 = (–1)3 (11) = –11 A13 = (–1)1+3 M13 = (–1)4 0 = 0 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 (1) = −1 A31 = (–1)3+1 M31 = (–1)4 (2) = 2 A32 = (–1)3+2 M32 = (–1)5 (–8) = 8 A33 = (–1)3+3 M33 = (–1)6 (3) = 3 Thus adj (A) = [■8(A11&A21&[email protected]&A22&[email protected]&A23&A33)] = [■8(0&3&2@−11&1&[email protected]&−1&3)] Calculating A (adj A) [■8(1&−1&[email protected]&0&−[email protected]&0&3)] [■8(0&3&2@−11&1&[email protected]&−1&3)] = [■8(1(0)−1(⤶7−11)+2(0)&1(3)−1(1)+2(−1)&1(2)−1(8)+2(3)@3(0)+0(⤶7−11)+(−2)(0)&3(3)+0(1)+(−2)(−1)&3(2)+0(8)+(−2)(3)@1(0)+0(⤶7−11)+3(0)&1(3)+0(1)+3(−1)&1(2)+0(8)+3(3))] = [■8(0+11+0&3−1−2&2−[email protected]−0−0&9+0+2&6+0−[email protected]−0+0&3+0−3&2+0+9)] = [■8(11&0&[email protected]&11&[email protected]&0&11)] = 11 [■8(1&0&[email protected]&1&[email protected]&0&1)] = 11I Calculating (adj A)A [■8(0&3&2@−11&1&[email protected]&−1&3)] [■8(1&−1&[email protected]&0&−[email protected]&0&3)] = [■8(0(1)+3(3)+2(1)&0(−1)+3(0)+2(0)&0(2)+3(−2)+2(3)@−11(1)+1(3)+8(1)&−11(−1)+1(0)+8(0)&−11(2)+1(−2)+8(3)@0(1)+(−1)(3)+3(1)&0(−1)+(−1)(0)+3(0)&0(2)+(−1)(−2)+3(3))] = [■8(0+9+2&−0+0+0&0−6+6@−11+3+8&11+0+0&−22−[email protected]−3+3&−0−0+0&−0+2+9)]a = [■8(11&0&[email protected]&11&[email protected]&0&11)] = 11 [■8(1&0&[email protected]&1&[email protected]&0&1)] = 11I Calculating |A| I |A|I = 11I Thus, A (adj(A)) = (adj A) A = |A| I = 11I Hence Proved

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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.