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प्रश्न
Solve the following equations by method of inversion :
4x – 3y – 2 = 0, 3x – 4y + 6 = 0
उत्तर
Matrix form of the given system of equations is
`[(4, -3),(3, -4)][(x),(y)] = [(2),(-6)]`
This of the form AX = B, where
A = `[(4, -3),(3, -4)], "X" = [(x),(y)] "and B" = [(2),(-6)]`
To determine X, we have to find A–1.
|A| = `|(4, -3),(3, -4)|`
= – 16 + 9
= – 7 ≠ 0
∴ A–1 exists
Consider AA–1 = I
`[(4, -3),(3, -4)] "A"^-1 = [(1, 0),(0, 1)]`
Applying R1 → R1 – R2, we get
`[(1, 1),(3, -4)] "A"^-1 = [(1, -1),(0, 1)]`
Applying R2 → R2 – 3R1, we get
`[(1, 1),(0, -7)] "A"^-1 = [(1, -1),(-3, 4)]`
Applying R2 → `(-1/7)`R2, we get
`[(1, 1),(0, 1)] "A"^-1 = [(1, -1),(3/7, (-4)/7)]`
Applying R1 → R1 – R2, we get
`[(1, 0),(0, 1)] "A"^-1 = [(4/7, (-3)/7),(3/7, (-4)/7)]`
∴ A–1 = `(1)/(7)[(4, -3),(3, -4)]`
Pre-multiplyig AX = by A–1, we get
A–1(AX) = A–1B
∴ (A–1A) X = A–1B
∴ IX = A–1B
∴ X = A–1B
∴ X = `(1)/(7)[(4, -3),(3, -4)][(2),(-6)]`
∴ `[(x),(y)] = (1)/(7)[(8 + 18),(6 + 24)]`
= `(1)/(7)[(26),(30)]`
= `[(26/7),(30/7)]`
∴ By equality of martices, we get
x = `(26)/(7)` and y = `(30)/(7)`.
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| April 2016 (in ₹.) | |||
| Rice | Wheat | Groundnut | |
| Shantaram | 15000 | 13000 | 12000 |
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| May 2016 (in ₹.) | |||
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Therefore by given conditions,
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R1 ↔ R2
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y + 2z = 24 ......(ii)
–z = ( ) ......(iii)
From (iii), we get, z = ( )
From (ii), we get, y = ( )
From (i), we get, x = ( )
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