Question

If A = `[(1, 0, -1),(2, 1, 3 ),(0, 1, 1)]`, then verify that A² + A = A(A + I), where I is 3 × 3 unit matrix.

Answer 1

We have, A = `[(1, 0, -1),(2, 1, 3 ),(0, 1, 1)]`

∴ A² = A · A

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

= `[(1 + 0 + 0, 0 + 0 - 1, -1 + 0 - 1),(2 + 2 + 0, 0 + 1 + 3, -2 + 3 + 3),(0 + 2 + 0, 0 + 1 + 1, 0 + 3 + 1)]`

= `[(1, -1, -2),(4, 4, 4),(2, 2, 4)]`

∴ A² + A = `[(1, -1, -2),(4, 4, 4),(2, 2, 4)] + [(1, 0, -4),(2, 1, 3),(0, 1, 1)]`

= `[(2, -1, -3),(6, 5, 7),(2, 3, 5)]`  ......(i)

Now, A + I = `[(1, 0, -1),(2, 1, 3),(0, 1, 1)] + [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

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

So, A(A + I) = `[(1, 0, -1),(2, 1, 3),(0, 1, 1)] [(2, 0, -1),(2, 2, 3),(0, 1, 2)]`

= `[(2 + 0 + 0, 0 + 0 - 1, -1 + 0 - 2),(4 + 2 + 0, 0 + 2 + 3, -2 + 3 + 6),(0 + 2 + 0, 0 + 2 + 1, 0 + 3 + 2)]`

= `[(2, -1, -3),(6, 5, 7),(2, 3, 5)]`  .....(iii)

From (i) and (ii)

We get A² + A = A(A + I)