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Question
If A = `[(4,5),(2,1)]`, show that `"A"^-1 = 1/6("A" - 5"I")`.
Solution
|A| = `|(4,5),(2,1)|` = 4 − 10 = − 6 ≠ 0
∴ A−1 exists.
Consider AA−1 = I
∴ `[(4,5),(2,1)] "A"^-1 = [(1,0),(0,1)]`
By `(1/4)"R"_1`, we get,
`[(1,5/4),(2,1)] "A"^-1 = [(1/4,0),(0,1)]`
By R2 − 2R1 we get,
`[(1,5/4),(0,-3/2)] "A"^-1 = [(1/4,0),(-1/2,1)]`
By `(- 2/3)"R"_2,` we get,
`[(1,5/4),(0,1)] "A"^-1 = [(1/4,0),(1/3,-2/3)]`
By `"R"_1 - 5/4 "R"_2,` we get,
`[(1,0),(0,1)] "A"^-1 = [(-1/6,5/6),(1/3,-2/3)]`
∴ A−1 = `1/6[(-1,5),(2,-4)]` ....(1)
`1/6("A" - 5"I") = 1/6{[(4,5),(2,1)] - 5 [(1,0),(0,1)]}`
`= 1/6 {[(4,5),(2,1)] - [(5,0),(0,5)]}`
`= 1/6 [(-1,5),(2,-4)]` ....(2)
From (1) and (2), we get `"A"^-1 = 1/6 ("A" - "5I")`
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