# Solve the following equations by the method of inversion: x + y+ z = 1, 2x + 3y + 2x = 2, ax + ay + 2az = 4, a ≠ 0. - Mathematics and Statistics

Sum

Solve the following equations by the method of inversion:

x + y+ z = 1, 2x + 3y + 2x = 2, ax + ay + 2az = 4, a ≠ 0.

#### Solution

The given equations can be written in the matrix form as:

[(1,1,1),(2,3,2),("a","a","2a")],[("x"),("y"),("z")] = [(1),(2),(4)]

This is of the form AX = B, where

A = [(1,1,1),(2,3,2),("a","a","2a")], "X" = [("x"),("y"),("z")]  "and"  "B" = [(1),(2),(4)]

Let us find A-1

|A| = [(1,1,1),(2,3,2),("a","a","2a")]

= 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)

= 4a - 2a - a

= a ≠ 0 ∴ A-1 exists.

Consider AA-1 = I

∴ [(1,1,1),(2,3,2),("a","a","2a")] "A"^-1 = [(1,0,0),(0,1,0),(0,0,1)]

By R2 - 2R1 and R3 - aR1, we get

[(1,1,1),(0,1,0),(0,0,"a")] = "A"^-1 = [(1,0,0),(-2,1,0),(-"a",0,1)]

By R1 - R2, we get

[(1,0,0),(0,1,0),(0,0,"a")] = "A"^-1 = [(3,-1,0),(-2,1,0),(-"a",0,1)]

By (1/"a")"R"_3, we get

[(1,0,0),(0,1,0),(0,0,1)] "A"^-1 = [(4,-1,-1/"a"),(-2,1,0),(-1,0,1/"a")]

∴ "A"^-1 = [(4,-1,-1/"a"),(-2,1,0),(-1,0,1/"a")]

ow, premultiply AX = B by A-1, we get,

A-1(AX) = A-1B

∴ (A-1A)X = A-1B

∴ IX = A-1B

∴ X = [(4,-1,-1/"a"),(-2,1,0),(-1,0,1/"a")][(1),(2),(4)]

∴ [("x"),("y"),("z")] = [(4-2-4/"a"),(-2+2+0),(-1+0+4/"a")] = [(2-4/"a"),(0),(4/"a" - 1)]

By equality of matrices,

x = 2 - 4/"a" y = 0, z = 4/"a" - 1 is the required solution.

Concept: Application of Matrices
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