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प्रश्न
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`
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उत्तर
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 Aij is the cofactor of aij.
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.
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