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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`
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.
`A_11 = (-1)^{1 + 1}[(-6,5),(9,-20)] = (-1)^2[120 - 45]`
= `1 xx 75 = 75`
`A_12 = (-1)^{1 + 2}[(4,5),(6,-20)] = (-1)^3[-80 - 30]`
= `-1 xx (-110) = 110`
`A_13 = (-1)^{1 + 3}[(4,-6),(6,9)] = (-1)^4[36 + 36]`
= `1 xx 72 = 72`
`A_21 = (-1)^{2 + 1}[(3,10),(9,-20)] = (-1)^3[-60 - 90]`
= `-1 xx (-150) = 150`
`A_22 = (-1)^{2 + 2}[(2,10),(6,-20)] = (-1)^4[-40 - 60]`
= `1 xx (-100) = -100`
`A_23 = (-1)^{2 + 3}[(2,3),(6,9)] = (-1)^5[18 - 18] = 0`
`A_31 = (-1)^{3 + 1}[(3,10),(-6,5)] = (-1)^4[15 + 60]`
= `1 xx 75 = 75`
`A_32 = (-1)^{3 + 2}[(2,10),(4,5)] = (-1)^5[10 - 40]`
= `-1 xx (-30) = 30`
`A_33 = (-1)^{3 + 3}[(2,3),(4,-6)] = (-1)^6[-12 - 12]`
= `1 xx (-24) = -24`
∴ adj A = `[(75,110,72),(150,-100,0),(75,30,-24)]`
= `[(75,150,75),(110,-100,30),(72,0,-24)]`
|A| = `a_11A_11 + a_12A_12 + a_13A_13`
= `2 xx 75 + 3 xx 110 + 10 xx 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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