Find solution of equations- Equations given

Chapter 4 Class 12 Determinants
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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 π/π = u , π/π = v , & π/π = w The system of equations become 2u + 3v + 10w = 4 4u β 6v + 5w = 1 6u + 9v β 20w = 2 Writing equation as AX = B [β 8(2&3&[email protected]&β6&[email protected]&9&β20)] [β 8(π’@π£@π€)] = [β 8([email protected]@2)] Hence A = [β 8(2&3&[email protected]&β6&[email protected]&9&β20)] , X = [β 8(π’@π£@π€)] & B = [β 8([email protected]@2)] Calculating |A| |A| = |β 8(2&3&[email protected]&β6&[email protected]&9&β20)| = 2 |β 8(β6&[email protected]&β20)| β 3 |β 8(4&[email protected]&β20)| + 10 |β 8(4&β[email protected]&9)| = 2 (120 β 45) β3 (β80 β 30) + 10 ( 36 + 36) = 2 (75) β3 (β110) + 10 (72) = 150 + 330 + 720 = 1200 β΄ |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]&β6&[email protected]&9&β20)] M11 = |β 8(β6&[email protected]&β20)| = 120 β 45 = 75 M12 = |β 8(4&[email protected]&β20)| = (β80 β 30) = β110 M13 = |β 8(4&β[email protected]&9)| = 36 β36 = 72 M21 = |β 8(3&[email protected]&β20)| = β60 β 90 = β150 M22 = |β 8(2&[email protected]&β20)| = β40 β 60 = β100 M23 = |β 8(2&[email protected]&9)| = 18 β 18 = 0 M31 = |β 8(3&10@β6&5)| = 15 + 60 = 75 M32 = |β 8(2&[email protected]&5)| = 10 β 40 = β30 M33 = |β 8(2&[email protected]&β6)| = β12 β 12 = β24 Now, A11 = γ"(β1)" γ^(1+1) M11 = (β1)2 . 75 = 75 A12 = γ"(β1)" γ^"1+2" M12 = γ"(β1)" γ^3 . (β110) = 110 A13 = γ(β1)γ^(1+3) M13 = γ(β1)γ^4 . (72) = 72 A21 = γ(β1)γ^(2+1) M21 = γ(β1)γ^3 . (β150) = 150 A22 = γ(β1)γ^(2+2) M22 = (β1)4 . (β100) = β100 A23 = γ(β1)γ^(2+3). M23 = γ(β1)γ^5. 0 = 0 A31 = γ(β1)γ^(3+1). M31 = γ(β1)γ^4 . 75 = 75 A32 = γ(β1)γ^(3+2) . M32 = γ(β1)γ^5. (β30) = 30 A33 = γ(β1)γ^(3+3) . M33 = (β1)6 . β24 = β24 Thus, adj A = [β 8(75&150&[email protected]&β110&[email protected]&0&β24)] Now, A-1 = 1/(|A|) adj A A-1 = 1/1200 [β 8(75&150&[email protected]&β110&[email protected]&0&β24)] Also, X = Aβ1 B Putting Values [β 8(π’@π£@π€)]= 1/1200 [β 8(75&150&[email protected]&β110&[email protected]&0&β24)] [β 8([email protected]@2)] [β 8(π’@π£@π€)]= 1/1200 [β 8(75(4)+150(1)+75(4)@110(4)+(β110)(1)+30(1)@72(4)+0(1)+(β24)2)] [β 8(π’@π£@π€)] = 1/1200 [β 8([email protected]β[email protected]+0β48)] = 1/1200 [β 8([email protected]@140)] [β 8(π’@π£@π€)] = [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 = π/π 1/2 = 1/π₯ x = 2 Putting v = π/π 1/3 = 1/π¦ y = 3 Putting w = π/π 1/5 = 1/π§ z = 5