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