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Question
If A = `[(5, 3),(-1, -2)]`, show that A2 – 3A – 7I2 = O2. Hence find A–1
Solution
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)]`
– 3A = `-3[(5, 3),(-1, -2)]`
= `[(-15, -9),(3, 6)]`
– 7I2 = `-7[(1, 0),(0, 1)]`
= `[(-7, 0),(0, -7)]`
A2 – 3A – 7I2
= `[(22, 9),(-3, 1)] + [(15, -9),(3, 6)] + [(-7, 0),(0, -7)]`
= `[(0, 0),(0, 0)]`
∴ A2 – 3A – 7I2 = O2
Post multiply this equation by A–1
A2A–1 – 3A A–1 – 7I2 A–1 = 0
A – 3I – 7A–1 = 0
A – 3I = 7 A–1
A–1 = `1/7 ("A" - 3"I")`
= `1/7 [((5, 3),(-1, -2)), -3((1, 0),(0, 1))]`
= `1/7 [((5, 3),(-1, -2)) + ((-3, 0),(0, -3))]`
A–1 = `1/7[(2, 3),(-1, -5)]`
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