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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.5, 4 Verify A (adj A) = (adj A) A = |๐ด|I for A = [โ– 8(1&โˆ’1&2@3&0&โˆ’2@1&0&3)] Calculating |๐ด| |A| = |โ– 8(1&โˆ’1&2@3&0&โˆ’2@1&0&3)| = 1 |โ– 8(0&โˆ’2@0&3)| โ€“ ( โ€“ 1) |โ– 8(3&โˆ’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&โˆ’1&2@3&0&โˆ’2@1&0&3)] M11 = |โ– 8(0&โˆ’2@0&3)| = 0(3) โ€“ 0(โ€“2) = 0 M12 = |โ– 8(3&โˆ’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(โˆ’1&2@0&3)| = โˆ’1(3) โ€“ 0(2) = โˆ’3 M22 = |โ– 8(1&2@1&3)| = 1(3) โ€“ 1(2) = 1 M23 = |โ– 8(1&โˆ’1@1&0)| = 1(0) โ€“ 1(โˆ’1) = 1 M31 = |โ– 8("-" 1&2@0&"-" 2)| = -1(โ€“2) โ€“ 0(2) = 2 M32 = |โ– 8(1&2@3&โˆ’2)| = 1(โ€“2) โ€“ 3(2) = โ€“8 M33 = |โ– 8(1&โˆ’1@3&0)| = 1(0) โ€“ 3(โˆ’1) = 3 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 (โˆ’3) = 3 A22 = ( โ€“ 1)2+2 M22 = ( โ€“ 1)4 . 1 = 1 A23 = ( โ€“ 1)2+3 M23 = ( โ€“ 1)5 ( 1) = โˆ’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@โˆ’11&1&8@0&โˆ’1&3)] Calculating A (adj A) [โ– 8(1&โˆ’1&2@3&0&โˆ’2@1&0&3)] [โ– 8(0&3&2@โˆ’11&1&8@0&โˆ’1&3)] = [โ– 8(1(0)โˆ’1(โคถ7โˆ’11)+2(0)&1(3)โˆ’1(1)+2(โˆ’1)&1(2)โˆ’1(8)+2(3)@3(0)+0(โคถ7โˆ’11)+(โˆ’2)(0)&3(3)+0(1)+(โˆ’2)(โˆ’1)&3(2)+0(8)+(โˆ’2)(3)@1(0)+0(โคถ7โˆ’11)+3(0)&1(3)+0(1)+3(โˆ’1)&1(2)+0(8)+3(3))] = [โ– 8(0+11+0&3โˆ’1โˆ’2&2โˆ’8+6@0โˆ’0โˆ’0&9+0+2&6+0โˆ’6@0โˆ’0+0&3+0โˆ’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@โˆ’11&1&8@0&โˆ’1&3)] [โ– 8(1&โˆ’1&2@3&0&โˆ’2@1&0&3)] = [โ– 8(0(1)+3(3)+2(1)&0(โˆ’1)+3(0)+2(0)&0(2)+3(โˆ’2)+2(3)@โˆ’11(1)+1(3)+8(1)&โˆ’11(โˆ’1)+1(0)+8(0)&โˆ’11(2)+1(โˆ’2)+8(3)@0(1)+(โˆ’1)(3)+3(1)&0(โˆ’1)+(โˆ’1)(0)+3(0)&0(2)+(โˆ’1)(โˆ’2)+3(3))] = [โ– 8(0+9+2&โˆ’0+0+0&0โˆ’6+6@โˆ’11+3+8&11+0+0&โˆ’22โˆ’2+24@0โˆ’3+3&โˆ’0โˆ’0+0&โˆ’0+2+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 |A| I |A|I = 11I Thus, A (adj(A)) = (adj A) A = |A| I = 11I โ‡’ A (adj(A)) = (adj A) A = |A| I Hence Proved

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