Question

Solve the system of linear equations using the matrix method.

x − y + z = 4

2x + y − 3z = 0

x + y + z = 2

Answer 1

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