Question

Find the inverse of the matrix `[(1, 1, 1),(1, 2, 3),(3, 2, 2)]` by elementary column transformation.

Answer 1

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

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

= 1(4 – 6) – 1(2 – 9) + 1(2 – 6)

= – 2 + 7 – 4

= 1 ≠ 0

∴ A is a non-singular square matrix.

∴ A^–1 exists.

We write A^–1A = I

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

Using `C_2 -> C_2 - C_1, C_3 -> C_3 - C_1`

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

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

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

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

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

∴  A^–1I = `[(-2, 0, 1),(7, -1, -2),(-4, 1, 1)]`

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