Ex 4.4

Chapter 4 Class 12 Determinants
Serial order wise

### Transcript

Ex 4.4, 4 Verify A (adj A) = (adj A) A = |๐ด|I for A = [โ 8(1&โ1&2@3&0&โ2@1&0&3)] Calculating |๐จ| |A| = |โ 8(1&โ1&2@3&0&โ2@1&0&3)| = 1 |โ 8(0&โ2@0&3)| โ (โ1) |โ 8(3&โ2@1&3)| +2 |โ 8(3&0@1&0)| = 1 (0 โ 0) + 1 (9 + 2) +2 (0 โ 0) = 11 Calculating adj A adj A = [โ 8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_23&A_33 )] A = [โ 8(1&โ1&2@3&0&โ2@1&0&3)] M11 = |โ 8(0&โ2@0&3)| = 0(3) โ 0(โ2) = 0 M12 = |โ 8(3&โ2@1&3)| = 3(3) โ 1(โ2) = 11 M13 = |โ 8(3&0@1&0)| = 3(0) โ 0(1) = 0 M21 = |โ 8(โ1&2@0&3)| = โ1(3) โ 0(2) = โ3 M22 = |โ 8(1&2@1&3)| = 1(3) โ 1(2) = 1 M23 = |โ 8(1&โ1@1&0)| = 1(0) โ 1(โ1) = 1 M31 = |โ 8("โ" 1&2@0&"โ" 2)| = -1(โ2) โ 0(2) = 2 M32 = |โ 8(1&2@3&โ2)| = 1(โ2) โ 3(2) = โ8 M33 = |โ 8(1&โ1@3&0)| = 1(0) โ 3(โ1) = 3 Now, A11 = (โ1)1 + 1 M11 = (โ1)2 0 = 0 A12 = (โ1)1+2 M12 = (โ1)3 (11) = โ11 A13 = (โ1)1+3 M13 = (โ1)4 0 = 0 A21 = (โ1)2+1 M21 = (โ1)3 (โ3) = 3 A22 = (โ1)2+2 M22 = (โ1)4 . 1 = 1 A23 = (โ1)2+3 M23 = (โ1)5 (1) = โ1 A31 = (โ1)3+1 M31 = (โ1)4 (2) = 2 A32 = (โ1)3+2 M32 = (โ1)5 (โ8) = 8 A33 = (โ1)3+3 M33 = (โ1)6 (3) = 3 Thus adj (A) = [โ 8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] = [โ 8(0&3&2@โ11&1&8@0&โ1&3)] Calculating A (adj A) [โ 8(1&โ1&2@3&0&โ2@1&0&3)] [โ 8(0&3&2@โ11&1&8@0&โ1&3)] = [โ 8(1(0)โ1(โคถ7โ11)+2(0)&1(3)โ1(1)+2(โ1)&1(2)โ1(8)+2(3)@3(0)+0(โคถ7โ11)+(โ2)(0)&3(3)+0(1)+(โ2)(โ1)&3(2)+0(8)+(โ2)(3)@1(0)+0(โคถ7โ11)+3(0)&1(3)+0(1)+3(โ1)&1(2)+0(8)+3(3))] = [โ 8(0+11+0&3โ1โ2&2โ8+6@0โ0โ0&9+0+2&6+0โ6@0โ0+0&3+0โ3&2+0+9)] = [โ 8(11&0&0@0&11&0@0&0&11)] = 11 [โ 8(1&0&0@0&1&0@0&0&1)] = 11I Calculating (adj A)A [โ 8(0&3&2@โ11&1&8@0&โ1&3)] [โ 8(1&โ1&2@3&0&โ2@1&0&3)] = [โ 8(0(1)+3(3)+2(1)&0(โ1)+3(0)+2(0)&0(2)+3(โ2)+2(3)@โ11(1)+1(3)+8(1)&โ11(โ1)+1(0)+8(0)&โ11(2)+1(โ2)+8(3)@0(1)+(โ1)(3)+3(1)&0(โ1)+(โ1)(0)+3(0)&0(2)+(โ1)(โ2)+3(3))] = [โ 8(0+9+2&โ0+0+0&0โ6+6@โ11+3+8&11+0+0&โ22โ2+24@0โ3+3&โ0โ0+0&โ0+2+9)]a = [โ 8(11&0&0@0&11&0@0&0&11)] = 11 [โ 8(1&0&0@0&1&0@0&0&1)] = 11I Calculating |A| I |A|I = 11I Thus, A (adj(A)) = (adj A) A = |A| I = 11I Hence Proved