Question

If A = `[(1, 2, 4),(4, 3, -2),(1, 0, -3)]`. Show that A^–1 exists and find A^–1 using column transformation.

Answer 1

We have

|A| = `|(1, 2, 4),(4, 3, -2),(1, 0, -3)|`

= 1(–9) – 2(– 12 + 2) + 4(– 3)

= – 9 + 20 – 12

= – 1 ≠ 0

∴ A^–1 exists.

We write

A^–1A = I

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

Using : `C_2 -> C_2 - 2C_1, C_3 -> 4C_1`

`A^-1[(1, 0, 0),(4, -5, -18),(1, -2, -7)] = [(1, -2, -4),(0, 1, 0),(0, 0, 1)]`

Using : `C_2 -> -4C_2 + C_3`

`A^-1[(1, 0, 0),(4, 2, -18),(1, 1, -7)] = [(1, 4, -4),(0, -4, 0),(0, 1, 1)]`

Using : `C_2 -> 1/2C_2`

`A^-1[(1, 0, 0),(4, 1, -18),(1, 1/2, -7)] = [(1, 2, -4),(0, -2, 0),(0, 1/2, 1)]`

Using : `C_1 -> C_1 - 4C_2, C_3 -> C_3 + 18C_2`

`A^-1[(1, 0, 0),(0, 1, 0),(-1, 1/2, 2)] = [(-7, 2, 32),(8, -2, -36),(-2, 1/2, 10)]`

Using : `C_3 -> 1/2C_3`

`A^-1[(1, 0, 0),(0, 1, 0),(-1, 1/2, 1)] = [(-7, 2, 16),(8, -2, -18),(-2, 1/2, 5)]`

Using : `C_1 -> C_1 + C_3, C_2 -> C_2 - 1/2C_3`

`A^-1[(1, 0, 0),(0, 1, 0),(0, 0, 1)] = [(9, -6, 16),(-10, 7, -18),(3, -2, 5)]`

∴ A^–1 = `[(9, -6, 16),(-10, 7, -18),(3, -2, 5)]`