# Ex 4.5, 4 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 22, 2020 by Teachoo

Ex 4.5

Ex 4.5, 1

Ex 4.5, 2

Ex 4.5, 3 Important

Ex 4.5, 4 Important You are here

Ex 4.5, 5

Ex 4.5, 6 Important

Ex 4.5, 7

Ex 4.5, 8

Ex 4.5, 9

Ex 4.5, 10 Important

Ex 4.5, 11 Important

Ex 4.5, 12

Ex 4.5, 13

Ex 4.5, 14 Important

Ex 4.5, 15 Important

Ex 4.5, 16

Ex 4.5, 17 (MCQ) Important

Ex 4.5, 18 (MCQ) Important

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at Jan. 22, 2020 by Teachoo

Ex 4.5, 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&[email protected]−11&1&[email protected]&−1&3)] Calculating A (adj A) [■8(1&−1&[email protected]&0&−[email protected]&0&3)] [■8(0&3&[email protected]−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&[email protected]−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−[email protected]−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