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Misc 16 - Solve equations 2/x +3/y +10/z = 4 4/x + 6/y +5/z = 1

Misc. 16 - Chapter 4 Class 12 Determinants - Part 2
Misc. 16 - Chapter 4 Class 12 Determinants - Part 3 Misc. 16 - Chapter 4 Class 12 Determinants - Part 4 Misc. 16 - Chapter 4 Class 12 Determinants - Part 5 Misc. 16 - Chapter 4 Class 12 Determinants - Part 6 Misc. 16 - Chapter 4 Class 12 Determinants - Part 7 Misc. 16 - Chapter 4 Class 12 Determinants - Part 8

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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 𝟏/𝒙 = u , 𝟏/π’š = v , & 𝟏/𝒛 = w The system of equations become 2u + 3v + 10w = 4 4u – 6v + 5w = 1 6u + 9v – 20w = 2 Writing equation as AX = B [β– 8(2&3&[email protected]&βˆ’6&[email protected]&9&βˆ’20)] [β– 8(𝑒@𝑣@𝑀)] = [β– 8([email protected]@2)] Hence A = [β– 8(2&3&[email protected]&βˆ’6&[email protected]&9&βˆ’20)] , X = [β– 8(𝑒@𝑣@𝑀)] & B = [β– 8([email protected]@2)] Calculating |A| |A| = |β– 8(2&3&[email protected]&βˆ’6&[email protected]&9&βˆ’20)| = 2 |β– 8(βˆ’6&[email protected]&βˆ’20)| – 3 |β– 8(4&[email protected]&βˆ’20)| + 10 |β– 8(4&βˆ’[email protected]&9)| = 2 (120 – 45) –3 (–80 – 30) + 10 ( 36 + 36) = 2 (75) –3 (–110) + 10 (72) = 150 + 330 + 720 = 1200 ∴ |A|β‰  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) = [β– 8(A11&A12&[email protected]&A22&[email protected]&A32&A33)]^β€² = [β– 8(A11&A21&[email protected]&A22&[email protected]&A23&A33)] A = [β– 8(2&3&[email protected]&βˆ’6&[email protected]&9&βˆ’20)] M11 = |β– 8(βˆ’6&[email protected]&βˆ’20)| = 120 – 45 = 75 M12 = |β– 8(4&[email protected]&βˆ’20)| = (–80 – 30) = –110 M13 = |β– 8(4&βˆ’[email protected]&9)| = 36 –36 = 72 M21 = |β– 8(3&[email protected]&βˆ’20)| = βˆ’60 – 90 = –150 M22 = |β– 8(2&[email protected]&βˆ’20)| = –40 – 60 = –100 M23 = |β– 8(2&[email protected]&9)| = 18 – 18 = 0 M31 = |β– 8(3&10@βˆ’6&5)| = 15 + 60 = 75 M32 = |β– 8(2&[email protected]&5)| = 10 – 40 = –30 M33 = |β– 8(2&[email protected]&βˆ’6)| = –12 – 12 = –24 Now, A11 = γ€–"(–1)" γ€—^(1+1) M11 = (–1)2 . 75 = 75 A12 = γ€–"(–1)" γ€—^"1+2" M12 = γ€–"(–1)" γ€—^3 . (–110) = 110 A13 = γ€–(βˆ’1)γ€—^(1+3) M13 = γ€–(βˆ’1)γ€—^4 . (72) = 72 A21 = γ€–(βˆ’1)γ€—^(2+1) M21 = γ€–(βˆ’1)γ€—^3 . (–150) = 150 A22 = γ€–(βˆ’1)γ€—^(2+2) M22 = (–1)4 . (–100) = –100 A23 = γ€–(βˆ’1)γ€—^(2+3). M23 = γ€–(βˆ’1)γ€—^5. 0 = 0 A31 = γ€–(βˆ’1)γ€—^(3+1). M31 = γ€–(βˆ’1)γ€—^4 . 75 = 75 A32 = γ€–(βˆ’1)γ€—^(3+2) . M32 = γ€–(βˆ’1)γ€—^5. (–30) = 30 A33 = γ€–(βˆ’1)γ€—^(3+3) . M33 = (–1)6 . –24 = –24 Thus, adj A = [β– 8(75&150&[email protected]&βˆ’110&[email protected]&0&βˆ’24)] Now, A-1 = 1/(|A|) adj A A-1 = 1/1200 [β– 8(75&150&[email protected]&βˆ’110&[email protected]&0&βˆ’24)] Also, X = Aβˆ’1 B Putting Values [β– 8(𝑒@𝑣@𝑀)]= 1/1200 [β– 8(75&150&[email protected]&βˆ’110&[email protected]&0&βˆ’24)] [β– 8([email protected]@2)] [β– 8(𝑒@𝑣@𝑀)]= 1/1200 [β– 8(75(4)+150(1)+75(4)@110(4)+(βˆ’110)(1)+30(1)@72(4)+0(1)+(βˆ’24)2)] [β– 8(𝑒@𝑣@𝑀)] = 1/1200 [β– 8([email protected]βˆ’[email protected]+0βˆ’48)] = 1/1200 [β– 8([email protected]@140)] [β– 8(𝑒@𝑣@𝑀)] = [β– 8(1/[email protected]/[email protected]/5)] Hence u = 1/2 , v = 1/3 , & w = 1/5 Thus, x = 2, y = 3 & z = 5 Putting u = 𝟏/𝒙 1/2 = 1/π‘₯ x = 2 Putting v = 𝟏/π’š 1/3 = 1/𝑦 y = 3 Putting w = 𝟏/𝒛 1/5 = 1/𝑧 z = 5

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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.