Question

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

Answer 1

Matrix m the given system  of equations is

`[(2, -1, 1),(1, 2, 3),(3, 1, -4)][(x),(y),(z)] = [(1),(8),(1)]`

This is of the form AX = B

where A = `[(2, -1, 1),(1, 2, 3),(3, 1, -4)], "X" = [(x),(y),(y),(z)]`

and B = `[(1),(8),(1)]`

To determine X , we have to find A^–1

Now, |A| = `|(2, -1, 1),(1, 2, 3),(3, 1, -4)|`

= 2(– 8 – 3) + 1(–4 – 9) + 1(1 – 6)
= – 22 – 13 – 5
= – 40 ≠ 0
∴ A^–1 exists.
Consider AA^–1 = I

∴ `[(2, -1, 1),(1, 2, 3),(3, 1, -4)] "A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Applying R₁ ↔ R₂, we get

`[(1, 2, 3),(2, -1, 1),(3, 1, -4)] "A"^-1 = [(0, 1, 0),(1, 0, 0),(0, 0, 1)]`

Applying R₂ → R₂ – 2R₁, R₃ → R₃ – 3R₁, we get

`[(1, 2, 3),(0, -5, -5),(0, -5, -13)] "A"^-1 = [(0, 1, 0),(1, -2, 0),(0, -3, 1)]`

Applying R₂ → `((-1)/5)` R₂, we get

`[(1, 2, 3),(0, 1,1),(0, -5, -13)] "A"^-1 = [(0, 1, 0),((-1)/5, 2/5, 0),(0, -3, 1)]`

Appying R₁ → R₁ – 2R₂, R₃ → R₃ – 5R₂, we get

`[(1, 0, 1),(0, 1, 1),(0, 0, -8)] "A"^-1 = [(2/5, 1/5, 0),((-1)/5, 2/5, 0),(-1, -1, 1)]`

Applying R₃ → `((-1)/8)` R₃, we get

`[(1, 0, 0),(0, 1, 1),(0, 0, 1)] "A"^-1 = [(2/5, 1/5, 0),((-1)/5, 2/5, 0),(1/8, 1/8, (-1)/8)]`

Appying R₁ → R₁ – R₃, R₂ → R₂ – R₃, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] "A"^-1 = [(11/40, 3/40, 1/8),((-13)/40, 11/40, 1/8),(1/8, 1/8, (-1)/8)]`

∴ A^–1 = `[(11/40, 3/40, 1/8),((-13)/40, 11/40, 1/8),(1/8, 1/8, (-1)/8)]`

∴ A^–1 = `(1)/(40)[(11, 3, 5),(-13, 11, 5),(5, 5, -5)]`

Pre-mutiplying AX = B by A^–1, we get
A^–1(AX) = A^–1B
∴ (A^–1 A)X = A^–1 B
∴ Ix = A^–1 B
∴ X = A^–1 B                  ...(i)

∴ X = `(1)/(40) = [(11, 3, 5),(-13, 11, 5),(5, 5, -5)] [(1),(8),(1)]`         ...[From (i)]

∴ `(1)/(40)[(11 + 4 + 5),(-13 + 88 + 5),(5 + 40 - 5)]`

= `(1)/(40)[(40),(80),(40)]`

∴ `[(x),(y),(z)] = [(1),(2),(3)]`

∴ By equality of matrices, we get
x = 1, y = 2, z = 1.