Question

(
c)
Solve the following equations by matrix inversion method
x+y+z=3
3x - 2y + 3z = 4
5x + 5y +z= 11​

Answer 1

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

`[(1,1,1),(0,1,1),(1,1,-1)][("x"),("y"),("z")] = [(-1),(2),(3)]`

This is of the form AX = B, where

A = `[(1,1,1),(0,1,1),(1,1,-1)], "X" = [("x"),("y"),("z")], "B" = [(-1),(2),(3)]`

Let us find A-1.

|A| = `|(1,1,1),(0,1,1),(1,1,-1)|`

= 1(- 1 - 1) - 1(0 - 1) + 1(0 - 1)

= - 2 + 1 - 1

= - 2 ≠ 0

∴ A-1 exists.

Consider AA-1 = I

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

By R3 - R1, we get

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

By `"R"_1 - "R"_2`, we get,

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

By `(- 1/2)"R"_3`, we get

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

By `"R"_2 - "R"_3`, we get

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

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

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

A-1(AX) = A-1B

∴ (A-1A)X = A-1B

∴ IX = A-1B

∴ X = `[(-1-2+0),(1/2+2+3/2),(-1/2+0-3/2)]`

∴ `[("x"),("y"),("z")] = [(-3),(4),(-2)]`

∴ by equality of the matrices, x = -3, y = 4, z = - 2 is the required solution..