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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 4.4, 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‘จ| |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("โˆ’" 1&2@0&"โˆ’" 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 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 (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)]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

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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.