Question

Solve the following equations by the method of inversion:

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

Answer 1

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

By R₁ - R₂, we get

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

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

`[(1,0,),(0,1,),(0,0,)] "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.