Solve the following equations by the method of inversion.
2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution
A = `[(2, -1, 1),(1, 2, 3),(3, 1, -4)], "B" = [(1),(8),(1)]`
x = `[(x),(y),(z)]`
|A| = 2(-8 - 3) + 1(-4 - 9) + 1(1 - 6)
= 2(-11) + 1(-13) + 1(-5)
= -22 - 13 - 5
|A| = -40 ≠ 0
∴ A-1 exist.
M11 = `|(2,3), (1,-4)| = -8 - 3 = -11`
M12 = `|(1,3),(3,-4)| = -4 - 9 = -13`
M13 = `|(1,2),(3,1)| = 1 - 6 = -5`
M21 = `|(-1,1),(1,-4)| = 4 - 1 = 3`
M22 = `|(2,1),(3,-4)| = -8 - 3 = -11`
M23 = `|(2,-1),(3,1)| = 2 + 3 = 5`
M31 = `|(-1,1),(2,3)| = -3 - 2 = -5`
M32 = `|(2,1),(1,3)| = 6 - 1 = 5`
M33 = `|(2,-1),(1,2)| = 4 + 1 = 5`
Aij = (-1)i + j.Mij
A11 = -11, A12 = 13, A13 = -5
A21 = -3, A22 = -11, A23 = -5
A31 = -5, A32 = -5, A33 = 5
cofactor matrix = `[(-11,13,-5),(-3,-11,-5),(-5,-5,5)]`
Adjoint Method = `[(-11,-3,-5),(13,-11,-5),(-5,-5,5)]`
`A^{-1} = 1/|A|`(Adjoint Method)
`A^{-1} = (-1)/40[(-11,-3,-5),(13,-11,-5),(-5,-5,5)]`
By method of Inversion,
X = A-1.B
= `(-1)/40[(-11,-3,-5),(13,-11,-5),(-5,-5,5)][(1),(8),(1)]`
= `(-1)/40[(-11 - 24 - 5),(13 - 88 - 5),(-5 - 40 + 5)]`
= `(-1)/40[(-40),(-80),(-40)] = [(1),(2),(1)]`
∴ x = 1, y = 2, z = 1