Solve the following equations by method of inversion.

2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1

#### Solution

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 R_{1} ↔ R_{2}, we get

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

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

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

Applying R_{2} → `((-1)/5)` R_{2}, 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 R_{1} → R_{1 }– 2R_{2}, R_{3} → R_{3} – 5R_{2}, 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 R_{3} → `((-1)/8)` R_{3}, 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 R_{1} → R_{1 }– R_{3}, R_{2} → R_{2} – R_{3}, 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^{–1}B

∴ (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.