Question

Solve the system of linear equations using the matrix method.

2x + y + z = 1

x – 2y – z = `3/2`

3y – 5z = 9

Answer 1

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`