# Solve the following equations by method of inversion.2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1 - Mathematics and Statistics

Sum

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

Concept: Application of Matrices
Is there an error in this question or solution?
Chapter 2: Matrices - Exercise 2.6 [Page 79]

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