Check sibling questions


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Misc 7 Solve the system of the following equations 2/x + 3/y + 10/z = 4 4/x + 6/y + 5/z = 1 6/x + 9/y + 20/z = 2 The system of equations are 2/x + 3/y + 10/z = 4 4/x + 6/y + 5/z = 1 6/x + 9/y + 20/z = 2 Now let ๐Ÿ/๐’™ = u , ๐Ÿ/๐’š = v , & ๐Ÿ/๐’› = w The system of equations become 2u + 3v + 10w = 4 4u โ€“ 6v + 5w = 1 6u + 9v โ€“ 20w = 2 Writing equation as AX = B [โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)] [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)] = [โ– 8(4@1@2)] Hence A = [โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)] , X = [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)] & B = [โ– 8(4@1@2)] Calculating |A| |A| = |โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)| = 2 |โ– 8(โˆ’6&5@9&โˆ’20)| โ€“ 3 |โ– 8(4&5@6&โˆ’20)| + 10 |โ– 8(4&โˆ’6@6&9)| = 2 (120 โ€“ 45) โ€“3 (โ€“80 โ€“ 30) + 10 ( 36 + 36) = 2 (75) โ€“3 (โ€“110) + 10 (72) = 150 + 330 + 720 = 1200 โˆด |A|โ‰  0 So, the system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj (A) = [โ– 8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^โ€ฒ = [โ– 8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [โ– 8(2&3&10@4&โˆ’6&5@6&9&โˆ’20)] M11 = |โ– 8(โˆ’6&5@9&โˆ’20)| = 120 โ€“ 45 = 75 M12 = |โ– 8(4&5@6&โˆ’20)| = (โ€“80 โ€“ 30) = โ€“110 M13 = |โ– 8(4&โˆ’6@6&9)| = 36 โ€“36 = 72 M21 = |โ– 8(3&10@9&โˆ’20)| = โˆ’60 โ€“ 90 = โ€“150 M22 = |โ– 8(2&10@6&โˆ’20)| = โ€“40 โ€“ 60 = โ€“100 M23 = |โ– 8(2&3@6&9)| = 18 โ€“ 18 = 0 M31 = |โ– 8(3&10@โˆ’6&5)| = 15 + 60 = 75 M32 = |โ– 8(2&10@4&5)| = 10 โ€“ 40 = โ€“30 M33 = |โ– 8(2&3@4&โˆ’6)| = โ€“12 โ€“ 12 = โ€“24 Now, A11 = ใ€–"(โ€“1)" ใ€—^(1+1) M11 = (โ€“1)2 . 75 = 75 A12 = ใ€–"(โ€“1)" ใ€—^"1+2" M12 = ใ€–"(โ€“1)" ใ€—^3 . (โ€“110) = 110 A13 = ใ€–(โˆ’1)ใ€—^(1+3) M13 = ใ€–(โˆ’1)ใ€—^4 . (72) = 72 A21 = ใ€–(โˆ’1)ใ€—^(2+1) M21 = ใ€–(โˆ’1)ใ€—^3 . (โ€“150) = 150 A22 = ใ€–(โˆ’1)ใ€—^(2+2) M22 = (โ€“1)4 . (โ€“100) = โ€“100 A23 = ใ€–(โˆ’1)ใ€—^(2+3). M23 = ใ€–(โˆ’1)ใ€—^5. 0 = 0 A31 = ใ€–(โˆ’1)ใ€—^(3+1). M31 = ใ€–(โˆ’1)ใ€—^4 . 75 = 75 A32 = ใ€–(โˆ’1)ใ€—^(3+2) . M32 = ใ€–(โˆ’1)ใ€—^5. (โ€“30) = 30 A33 = ใ€–(โˆ’1)ใ€—^(3+3) . M33 = (โ€“1)6 . โ€“24 = โ€“24 Thus, adj A = [โ– 8(75&150&75@110&โˆ’110&30@72&0&โˆ’24)] Now, A-1 = 1/(|A|) adj A A-1 = ๐Ÿ/๐Ÿ๐Ÿ๐ŸŽ๐ŸŽ [โ– 8(๐Ÿ•๐Ÿ“&๐Ÿ๐Ÿ“๐ŸŽ&๐Ÿ•๐Ÿ“@๐Ÿ๐Ÿ๐ŸŽ&โˆ’๐Ÿ๐Ÿ๐ŸŽ&๐Ÿ‘๐ŸŽ@๐Ÿ•๐Ÿ&๐ŸŽ&โˆ’๐Ÿ๐Ÿ’)] Also, X = Aโˆ’1 B Putting Values [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)]= 1/1200 [โ– 8(75&150&75@110&โˆ’110&30@72&0&โˆ’24)] [โ– 8(4@1@2)] [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)]= 1/1200 [โ– 8(75(4)+150(1)+75(4)@110(4)+(โˆ’110)(1)+30(1)@72(4)+0(1)+(โˆ’24)2)] [โ– 8(๐‘ข@๐‘ฃ@๐‘ค)] = 1/1200 [โ– 8(300+150+150@440โˆ’100+60@288+0โˆ’48)] = 1/1200 [โ– 8(600@400@140)] [โ– 8(๐’–@๐’—@๐’˜)] = [โ– 8(๐Ÿ/๐Ÿ@๐Ÿ/๐Ÿ‘@๐Ÿ/๐Ÿ“)] Hence u = 1/2 , v = 1/3 , & w = 1/5 Thus, x = 2, y = 3 & z = 5 Putting u = ๐Ÿ/๐’™ 1/2 = 1/๐‘ฅ x = 2 Putting v = ๐Ÿ/๐’š 1/3 = 1/๐‘ฆ y = 3 Putting w = ๐Ÿ/๐’› 1/5 = 1/๐‘ง z = 5

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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo