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प्रश्न
Solve the system of linear equations using the matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
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उत्तर
Given system of equations,
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
The system of equations can be written as AX = B so X = A−1B
A = `[(1,-1,1),(2,1,-3),(1,1,1)]`, X = `[(x),(y),(z)]`, B = `[(4),(0),(2)]`
∴ |A| = `|(1,-1,1),(2,1,-3),(1,1,1)|`
= 1[1 + 3] − (−1)[2 + 3] + 1[2 − 1]
= 1 × 4 + 1 × 5 + 1 × 1
= 4 + 5 + 1
= 10 ≠ 0
The cofactors of the elements of matrix A are as follows:
A11 = `(-1)^(1 + 1) |(1,-3),(1,1)|`
= (−1)2 [1 + 3]
= 1 × 4
= 4
A12 = `(-1)^(1 + 2) |(2,-3),(1,1)|`
= (−1)3 [2 + 3]
= −1 × 5
= −5
A13 = `(-1)^(1 + 3) |(2,1),(1,1)|`
= (−1)4 [2 − 1]
= 1 × 1
= 1
A21 = `(-1)^(2 + 1) |(-1,1),(1,1)|`
= (−1)3 [−1 − 1]
= −1 × (−2)
= 2
A22 = `(-1)^(2 + 2) |(1,1),(1,1)|`
= (−1)4 [1 − 1]
= 0
A23 = `(-1)^(2 + 3) |(1,-1),(1,1)|`
= (−1)5 [1 + 1]
= −1 × 2
= −2
A31 = `(-1)^ (3 + 1) |(-1,1),(1,-3)|`
= (−1)4 [3 − 1]
= 1 × 2
= 2
A32 = `(-1)^(3 + 2) |(1,1),(2,-3)|`
= (−1)5 [−3 − 2]
= −1 × (−5)
= 5
A33 = `(-1)^(3 + 3) |(1,-1),(2,1)|`
= (−1)6 [1 + 2]
= 1 × 3
= 3
Hence the matrix made up of the elements of cofactors = `[(4,-5,1),(2,0,-2),(2,5,3)]`
∴ adj A = `[(4,-5,1),(2,0,-2),(2,5,3)] = [(4,2,2),(-5,0,5),(1,-2,3)]`
A−1 = `1/|A|` (adj A)
= `1/10 [(4,2,2),(-5,0,5),(1,-2,3)]`
∴ X = A−1B
= `1/10 [(4,2,2),(-5,0,5),(1,-2,3)] [(4),(0),(2)]`
= `1/10 [(16 + 0 + 4),(-20 + 10),(4 + 6)]`
= `1/10 [(20),(-10),(10)]`
= `[(2),(-1),(1)]`
⇒ `[(x),(y),(z)] = [(2),(-1),(1)]`
⇒ x = 2, y = −1 and z = 1
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