Question

Show that A = `[(5, 3),(-1, -2)]` satisfies the equation A² – 3A – 7I = O and hence find A^–1.

Answer 1

Given that A = `[(5, 3),(-1, -2)]`

A² = `"A" * "A"`

= `[(5, 3),(-1, -2)][(5, 3),(-1, -2)]`

= `[(25 - 3, 15 - 6),(-5 + 2, -3 + 4)]`

= `[(22, 9),(-3, 1)]`

A² – 3A – 7I = O

L.H.S. `[(2, 9),(-3, 1)] -3[(5, 3),(-1, -2)] -7[(1, 0),(0, 1)]`

⇒ `[(22, 9),(-3, 1)] - [(15, 9),(-3, -6)] - [(7, 0),(0, 7)]`

⇒ `[(22 - 15 - 7, 9 - 9 - 0),(-3 + 3 - 0, 1 + 6 - 7)]`

⇒ `[(0, 0),(0, 0)]` R.H.S.

We are given A² – 3A – 7I = O

⇒ A^–1 [A² – 3A – 7I] = A^–1O  ....[Pre-multiplying both sides by A^–1]

⇒ A^–1A · A – 3A^–1 · A – 7A^–1 I = O  .....[A^–1O = O]

⇒ I · A – 3I – 7A–1 I = O

⇒ A – 3I – 7A^–1 = O

⇒ –7A^–1 = 3I – A

⇒ A^–1 = `1/(-7) [3"I" - "A"]`

⇒ A^–1 = `1/(-7) [3((1, 0),(0, 1)) - ((5, 3),(-1,-2))]`

= `1/(-7) [3((1, 0),(0, 1)) - ((5, 3),(-1,-2))]`

= `1(-7) [(3 - 5, 0 - 3),(0 + 1, 3 + 2)]`

= `1/(-7) [(-2, -3),(1, 5)]`

Hence, A^–1 = `- 1/7 [(-2, -3),(1, 5)]`