Question

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/x = 2`

Answer 1

The given equation,

`2/x + 3/y + 10/z = 4`

`4/x - 6/y + 5/z = 1`

`6/x + 9/y - 20/z = 2`

Let, `1/x` = u, `1/y` = v, `1/z` = w

∴ 2u + 3v + 10w = 4

4u − 6v + 5w = 1

6u + 9v − 20w = 2

This can be written as AX = B, where

A = `[(2,3,10),(4,-6,5),(6,9,-20)]`, X = `[(u),(v),(w)]`, B = `[(4),(1),(2)]`

The element A_ij is the cofactor of a_ij.

A₁₁ = `(-1)^(1 + 1)[(-6,5),(9,-20)]`

= (−1)² [120 − 45]

= 1 × 75

= 75

A₁₂ = `(-1)^(1 + 2)[(4,5),(6,-20)]`

= (−1)³ [−80 − 30]

= −1 × (−110)

= 110

A₁₃ = `(-1)^(1 + 3)[(4,-6),(6,9)]`

= (−1)⁴ [36 + 36]

= 1 × 72

= 72

A₂₁ = `(-1)^(2 + 1)[(3,10),(9,-20)]`

= (−1)³ [−60 − 90]

= −1 × (−150)

= 150

A₂₂ = `(-1)^(2 + 2)[(2,10),(6,-20)]`

= (−1)⁴ [−40 − 60]

= 1 × (−100)

= −100

A₂₃ = `(-1)^(2 + 3)[(2,3),(6,9)]`

= (−1)⁵ [18 - 18]

= 0

A₃₁ = `(-1)^(3 + 1)[(3,10),(-6,5)]`

= (−1)⁴ [15 + 60]

= 1 × 75

= 75

A₃₂ = `(-1)^(3 + 2)[(2,10),(4,5)]`

= (−1)⁵ [10 − 40]

= −1 × (−30)

= 30

A₃₃ = `(-1)^(3 + 3)[(2,3),(4,-6)]`

= (−1)⁶ [−12 − 12]

= 1 × (−24)

= −24

∴ adj A = `[(75,110,72),(150,-100,0),(75,30,-24)]`

= `[(75,150,75),(110,-100,30),(72,0,-24)]`

|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃

= 2 × 75 + 3 × 110 + 10 × 72

= 150 + 330 + 720

= 1200

A⁻¹ = `1/|A|` (adj A)

= `1/1200[(75,150,75),(110,-100,30),(72,0,-24)]`

X = A⁻¹B

= `1/1200[(75,150,75),(110,-100,30),(72,0,-24)][(4),(1),(2)]`

`[(u),(v),(w)] = 1/1200[(300 + 150 + 150),(440 - 100 + 60),(288 + 0 - 48)]`

= `1/12000 [(600),(400),(240)]`

= `[(1/2),(1/3),(1/5)]`

∴ u = `1/2`, v = `1/3`, w = `1/5`

⇒ x = `1/u` = 2, y = `1/v` = 3, z = `1/w` = 5

Hence, the solutions of the system of equations are x = 2, y = 3, z = 5.