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Question
For the matrix A = `[(1,1,1),(1,2,-3),(2,-1,3)]` show that A3 − 6A2 + 5A + 11 I = 0. Hence, find A−1.
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Solution
A = `[(1,1,1),(1,2,-3),(2,-1,3)]`
A2 = AA
= `[(1,1,1),(1,2,-3),(2,-1,3)] [(1,1,1),(1,2,-3),(2,-1,3)]`
= `[(1+1+2,1+2-1,1-3+3),(1+2-6,1+4+3,1-6-9),(2-1+6,2-2-3,2+3+9)]`
= `[(4,2,1),(-3,8,-14),(7,-3,14)]`
A3 = A2A
= `[(4,2,1),(-3,8,-14),(7,-3,14)] [(1,1,1),(1,2,-3),(2,-1,3)]`
= `[(4+2+2,4+4-1,4-6+3),(-3+8-28,-3+16+14,-3-24-42),(7-3+28,7-6-14,7+9+42)]`
= `[(8,7,1),(-23,27,-69),(32,-13,58)]`
L.H.S. = A3 − 6A2 + 5A + 11 I
= `[(8,7,1),(-23,27,-69),(32,-13,58)] - 6 [(4,2,1),(-3,8,-14),(7,-3,14)] + 5 [(1,1,1),(1,2,-3),(2,-1,3)] + 11 [(1,0,0),(0,1,0),(0,0,1)]`
= `[(8,7,1),(-23,27,-69),(32,-13,58)] - [(24,12,6),(-18,48,-84),(42,-18,84)] + [(5,5,5),(5,10,-15),(10,-5,15)] + [(11,0,0),(0,11,0),(0,0,11)]`
= `[(8 - 24 + 5 + 11, 7 - 12 + 5 + 0, 1 - 6 + 5 + 0),(-23 + 18 + 5 + 0, 27 - 48 + 10 + 11, -69 + 84 - 15 + 0),(32 - 42 + 10 + 0,-13 + 18 - 5 + 0, 58 - 84 + 15 + 11)]`
= `[(0,0,0),(0,0,0),(0,0,0)]`
= 0
Hence, A3 − 6A2 + 5A + 11 I = 0
Now, A3 − 6A2 + 5A + 11 I = 0
A3 − 6A2 + 5A = −11 I
A2AA−1 = 6AAA−1 + 5AA−1 = 11 I A−1
11A−1 = −A2 + 6A − 5I
= `[(-4,-2,-1),(3,-8,14),(-7,3,-14)] + 6 [(1,1,1),(1,2,-3),(2,-1,3)] - 5 [(1,0,0),(0,1,0),(0,0,1)]`
= `[(-4,-2,-1),(3,-8,14),(-7,3,-14)] + [(6,6,6),(6,12,-18),(12,-6,18)] - [(5,0,0),(0,5,0),(0,0,5)]`
= `[(-3,4,5),(9,-1,-4),(5,-3,-1)]`
A−1 = `1/11 [(-3,4,5),(9,-1,-4),(5,-3,-1)]`
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