Question

व्यस्त पद्धतीने (method on inversion) खालील समीकरणे सोडवा:

2x – y + z = 1

x + 2y + 3z = 8

3x + y – 4z = 1

Answer 1

A = `[(2, -1, 1),(1, 2, 3),(3, 1, -4)], B = [(1),(8),(1)]`

x = `[(x),(y),(z)]`

|A| = 2(−8 − 3) + 1(−4 − 9) + 1(1 − 6)

= 2(−11) + 1(−13) + 1(−5)

= −22 − 13 − 5

|A| = − 40 ≠ 0

∴ A^-1 अस्तित्वात आहे.

M₁₁ = `|(2,3), (1,-4)| = −8 − 3 = −11`

M₁₂ = `|(1,3),(3,-4)| = −4 − 9 = −13`

M₁₃ = `|(1,2),(3,1)| = 1 − 6 = −5`

M₂₁ = `|(−1,1),(1,−4)| = 4 − 1 = 3`

M₂₂ = `|(2,1),(3,−4)| = -8 - 3 = -11`

M₂₃ = `|(2,-1),(3,1)| = 2 + 3 = 5`

M₃₁ = `|(-1,1),(2,3)| = -3 - 2 = -5`

M₃₂ = `|(2,1),(1,3)| = 6 - 1 = 5`

M₃₃ = `|(2,-1),(1,2)| = 4 + 1 = 5`

`A_(ij) = (−1)^(i + j) *M_(ij)`

A₁₁ = −11, A₁₂ = 13, A₁₃ = −5

A₂₁ = −3, A₂₂ = −11, A₂₃ = −5

A₃₁ = −5, A₃₂ = −5, A₃₃ = 5

सहगुणक आव्यूह = `[(−11,13,-5),(-3,-11,-5),(-5,-5,5)]`

सह-आव्यूह पद्धत = `[(-11,-3,-5),(13,-11,-5),(-5,-5,5)]`

`A^{-1} = 1/|A|`(सह-आव्यूह पद्धत)

`A^{-1} = (-1)/40[(-11,-3,-5),(13,-11,-5),(-5,-5,5)]`

व्यस्त आव्यूह पद्धतीनुसार,

X = A^-1 . B

= `(-1)/40[(-11,-3,-5),(13,-11,-5),(-5,-5,5)][(1),(8),(1)]`

= `(-1)/40[(-11 - 24 - 5),(13 - 88 - 5),(-5 - 40 + 5)]`

= `(-1)/40[(-40),(-80),(-40)]`

= `[(1),(2),(1)]`

∴ x = 1, y = 2, z = 1