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Question
Let A = `[(2, 3),(-1, 2)]`. Then show that A2 – 4A + 7I = O. Using this result calculate A5 also.
Solution
We have A2 = `[(2, 3),(-1, 2)] [(2, 3),(-1, 2)] = [(1, 12),(-4, 1)]`
– 4A = `[(-8, -12),(4, -8)]` and 7I = `[(7, 0),(0, 7)]`
Therefore, A2 – 4A + 7I = `[(1 - 8 + 7, 12 - 12 + 0),(-4 + 4 + 0, 1 - 8 + 7)]`
= `[(0, 0),(0, 0)]`
= O
⇒ A2 – 4A + 7I
Thus A3 = A.A2 = A(4A – 7I)
= 4(4A – 7I) – 7A
= 16A – 28I – 7A = 9A – 28I
and so A5 = A3A2
= (9A – 28I) (4A – 7I)
= 36A2 – 63A – 112A + 196I
= 36(4A – 7I) – 175A + 196I
= – 31A – 56I
= `-3"I"[(2, 3),(-1, 2)] -56[(1, 0),(0, 1)]`
= `[(-118, -93),(31, -118)]`
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