**Solve the following equations by method of inversion :** x + y – z = 2, x – 2y + z = 3 and 2x – y – 3z = – 1

#### Solution

Matrix form of the given system of equations is

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

This is of the form AX = B, where

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

To determine X, we have to find A^{–1}.

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

= 1(6 + 1) – 1(–3 – 2) –1(–1 + 4)

= 1(7) –1(–5)–1(3)

= 7 + 5 – 3

= 9 ≠ 0

∴ A^{–1} exists.

Consider AA^{–1} =

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

Applying R_{2} → R_{2} – R_{1} and R_{3} → R_{3} – 2R_{1}, we get

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

Applying R_{2} → `((-1)/3)` R_{2}, we get

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

Applying R_{1} → R_{1} – R_{2} and R_{3} → R_{3} + 3R_{2}, we get

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

Applying R_{3} → `(-1/3)` R_{3}, we get

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

Applying R_{1} → R_{1} + `(-1/3)` R_{3} and R_{2} → R_{2} + `(2/3)` R_{3}, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] "A"^-1 = [(7/9, 4/9, -1/9),(5/9, -1/9, -2/9),(1/3, 1/3, -1/3)]`

∴ A^{–1} = `(1)/(9)[(7, 4, -1),(5, -1, -2),(3, 3, -3)]`

Pre-multiplying AX = B by A^{–1}, we get

A^{–1}(AX) = A^{–1}B

∴ (A^{–1}A) X = A^{–1}B

∴ IX = A^{–1}B

∴ X = A^{–1}B

∴ X = `(1)/(9)[(7, 4, -1),(5, -1, -2),(3, 3, -3)][(2),(3),(-1)]`

∴ `[(x),(y),(z)] = (1)/(9)[(14 + 12 + 1),(10 - 3 + 2),(6 + 9 + 3)]`

= `(1)/(9)[(27),(9),(18)]`

= `[(3),(1),(2)]`

∴ By equality of martices, we get

x = 3, y = 1 and z = 2.