Question

Solve the following equations by using inversion method.

x + y + z = −1, x − y + z = 2 and x + y − z = 3

Answer 1

Matrix form of the given system of equations is,

`[(1, 1, 1),(1, -1, 1),(1, 1, -1)] [(x),(y),(z)] = [(-1),(2),(3)]`

This is of the form AX = B,

where, A = `[(1, 1, 1),(1, -1, 1),(1, 1, -1)], "X" = [(x),(y),(z)], "B" = [(-1),(2),(3)]`

Pre-multiplying AX = B by A⁻¹, we get

A⁻¹(AX) = A⁻¹B

∴ (A⁻¹A)X = A⁻¹B

∴ IX = A⁻¹B

∴ X = A⁻¹B     .......(i)

 To determine X, we have to find A⁻¹ 

|A| = `|(1, 1, 1),(1, -1, 1),(1, 1, -1)|`

= 1(1 − 1) − 1(−1 − 1) + 1(1 + 1)

= 2 + 2

= 4 ≠ 0

∴ A⁻¹ exists.

A₁₁ = (−1)¹⁺¹ M₁₁ = `|(-1, 1),(1, -1)|` = 1 − 1 = 0

A₁₂ = (−1)¹⁺² M₁₂ = `-|(1, 1),(1, -1)|` = −(−1 − 1) = 2

A₁₃ = (−1)¹⁺³ M₁₃ = `|(1, 1),(1, -1)|` = 1 + 1 = 2

A₂₁ = (−1)²⁺¹ M₂₁ = `-|(1, 1),(1, -1)|` = − (−1 − 1) = 2

A₂₂ = (−1)²⁺² M₂₂ = `|(1, 1),(1, -1)|` = −1 − 1 = −2

A₂₃ = (−1)²⁺³ M₂₃ = `-|(1, 1),(1, 1)|` = − (1 − 1) = 0

A₃₁ = (−1)³⁺¹ M₃₁ = `|(1, 1),(-1, 1)|` = 1 + 1 = 2

A₃₂ = (−1)³⁺² M₃₂ = `-|(1, 1),(1, 1)|` = −(1 − 1) = 0

A₃₃ = (−1)³⁺³ M₃₃ = `|(1, 1),(1, -1)|` = −1 − 1 = −2

Hence, the matrix of cofactors is

`|("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)| = [(0, 2, 2),(2, -2, 0),(2, 0, -2)]`

∴ adj A = `[(0, 2, 2),(2, -2, 0),(2, 0, -2)]`

= `2[(0, 1, 1),(1, -1, 0),(1, 0, -1)]`

∴ A−1 = `1/|"A"|` (adj A)

= `1/4 xx 2[(0, 1, 1),(1, -1, 0),(1, 0, -1)]`

∴ A−1 = `1/2[(0, 1, 1),(1, -1, 0),(1, 0, -1)]`

∴ X  = `1/2[(0, 1, 1),(1, -1, 0),(1, 0, -1)] [(-1),(2),(3)]`  .......[From (i)]

= `1/2[(5),(-3),(-4)]`

∴ `[(x),(y),(z)] = [(5/2),(-3/2),(-2)]`

By equality of matrices, we get

x = `5/2`, y = `-3/2`, z = −2