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Question
Solve the system of linear equations using the matrix method.
2x + y + z = 1
x – 2y – z = `3/2`
3y – 5z = 9
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Solution
The given equation,
2x + y + z = 1
x – 2y – z = `3/2`
3y – 5z = 9
The equation can be written as a system so X = A−1B
Where, A = `[(2,1,1),(1,-2,-1),(0,3,-5)]`, X = `[(x),(y),(z)]` and B = `[(1),(3/2),(9)]`
∴ |A| = `|(2,1,1),(1,-2,-1),(0,3,-5)|`
= 2[10 + 3] − 1[−5 + 0] + 1[3 + 0]
= 2 × 13 − 1 × (−5) + 1 × 3
= 26 + 5 + 3
= 34 ≠ 0
The cofactors of the elements of matrix A are as follows:
A11 = `(-1)^(1 + 1) |(-2,-1),(3,-5)|`
= (−1)2 [10 + 3]
= 1 × 13
= 13
A12 = `(-1)^(1 + 2) |(1,-1),(0,-5)|`
= (−1)3 [−5 + 0]
= −1 × (−5)
= 5
A13 = `(- 1)^(1 + 3) |(1,-2),(0,3)|`
= (−1)4 [3 + 0]
= 1 × 3
= 3
A21 = `(-1)^(2 + 1) |(1,1),(3,-5)|`
= (−1)3 [−5 − 3]
= −1 × (−8)
= 8
A22 = `(-1)^(2+2) |(2,1), (0,-5)|`
= (−1)4 [−10 − 0]
= 1 × −10
= −10
A23 = `(-1)^(2 + 3) |(2,1),(0,3)|`
= (−1)5 [6 − 0]
= −1 × 6
= −6
A31 = `(-1)^(3 + 1) |(1,1),(-2,-1)|`
= (−1)4 [−1 + 2]
= 1 × 1
= 1
A32 = `(-1)^(3 + 2) |(2,1),(1,-1)|`
= (−1)5 [−2 − 1]
= −1 × (−3)
= 3
A33 = `(-1)^(3 + 3) |(2,1),(1,-2)|`
= (−1)6 [−4 − 1]
= 1 × (−5)
= −5
Hence, the matrix made up of the elements of the cofactors = `[(13,5,3),(8,-10,-6),(1,3,-5)]`
∴ adj A = `[(13,5,3),(8,-10,-6),(1,3,-5)] = [(13,8,1),(5,-10,3),(3,-6,-5)]`
A−1 = `1/|A|` (adj A)
= `1/34 [(13,8,1),(5,-10,3),(3,-6,-5)]`
∴ X = A−1B
= `1/34 [(13,8,1),(5,-10,3),(3,-6,-5)] [(1),(3/2),(9)]`
= `1/34 [(13 + 12 + 9),(5 - 15 + 27),(3 - 9 - 45)]`
= `1/34 [(34),(17),(-51)]`
= `[(34/34),(17/34),((-51)/34)]`
⇒ `[(x),(y),(z)] = [(1),(1/2),((-3)/2)]`
⇒ x = 1, y = `1/2` and z = `(-3)/2`
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