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