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प्रश्न
Solve the following systems of linear equations by Cramer’s rule:
`3/x - 4/y - 2/z - 1` = 0, `1/x + 2/y + 1/z - 2` = 0, `2/x - 5/y - 4/z + 1` = 0
उत्तर
`3/x - 4/y - 2/z - 1` = 0
`1/x + 2/y + 1/z - 2` = 0
`2/x - 5/y - 4/z + 1` = 0
Put a = `1/x`, b = `1/y, c = 1/z`
3a – 4b – 2c = 1 .........(1)
a + 2b + c = 2 ..........(2)
2a – 5b – 4c = – 1 ............(3)
Δ = `|(3, -4, -2),(1, 2, 1),(2, -5, -4)|`
= 3(– 8 + 5) + 4(– 4 – 2) – 2(– 5 – 4)
= 3(– 3) + 4(– 6) – 2(– 9)
= – 9 – 24 + 18
= – 15 ≠ 0
Δa = `|(1, -4, -2),(2, 2, 1),(-1, -5, -4)|`
= 1(– 8 + 5) + 4(– 8 + 1) – 2(– 10 + 2)
= 1(– 3) + 4(– 7) – 2(– 8)
= – 3 – 28 + 16
= – 15
Δb = `|(3, 1, -2),(1, 2, 1),(2, -1, -4)|`
= 3(– 8 + 1) – 1(– 4 – 2) – 2(– 1 – 4)
= 3(– 7) – 1(– 6) – 2(– 5)
= – 21 + 6 + 10
= – 5
Δc = `|(3, -4, 1),(1, 2, 2),(2, -5, -1)|`
= 3(– 2 + 10) + 4(– 1 – 4) + 1(– 5 – 4)
= 24 – 20 – 9
= – 5
a = `Delta_"a"/Delta = (- 15)/(- 15)` = 1
⇒ `1/x` = 1
⇒ x = 1
b = `Delta_"b"/Delta = (-5)/(-15) = 1/3`
⇒ `1/y = 1/3`
⇒ y = 3
c = `Delta_"c"/Delta = (-5)/(-15) = 1/3`
⇒ `1/z = 1/3`
⇒ z = 3
∴ x = 1, y = 3, z = 3
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