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Question
Find the inverse of the matrices (if it exists).
`[(1,0,0),(3,3,0),(5,2,-1)]`
Solution
A ` = [(1,0,0),(3,3,0),(5,2,-1)]`
|A| = `|(1,0,0),(3,3,0),(5,2,-1)|`
= - 1[- 3 - 0]
= 1 × (- 3)
= - 3
`"A"_11 = (- 1)^(1 + 1) |(3,0),(2,-1)| = (- 1)^2 [- 3 - 0]`
`= 1 xx (- 3) = - 3`
`"A"_12 = (- 1)^(1 + 2) |(3,0),(5,-1)| = (- 1)^3 [- 3 - 0]`
`= - 1 xx (- 3) = 3`
`"A"_13 = (- 1)^(1 + 3) |(3,3),(5,2)| = (- 1)^4 [6 - 15]`
`= 1 xx (- 9) = - 9`
`"A"_21 = (- 1)^(2 + 1) |(0,0),(2,-1)| = (- 1)^3 [0 - 0] = 0`
`"A"_22 = (- 1)^(2 + 2) |(1,0),(5,-1)| = (- 1)^4 [- 1 - 0]`
`= 1 xx (- 1) = - 1`
`"A"_23 = (- 1)^(2 + 3) |(1,0),(5,2)| = (- 1)^5 [2 - 0]`
`= - 1 xx 2 = - 2`
`"A"_31 = (- 1)^(3 + 1) |(0,0),(3,0)| = (- 1)^4 [0 - 0]` = 0
`"A"_32 = (- 1)^(3 + 2) |(1,0),(3,3)| = (- 1)^5 [0 - 0]` = 0
`"A"_33 = (- 1)^(3 + 3) |(1,0),(3,3)| = (- 1)^6 [3 - 0] = 1 xx 3 = 3`
∴ adj A = `[(-3,3,-9),(0,-1,-2),(0,0,3)] = [(-3,0,0),(3,-1,0),(-9,-2,3)]`
`"A"^-1 = 1/abs "A" ("adjA")`
`= 1/abs "A" [("A"_11,"A"_21,"A"_31),("A"_12,"A"_22,"A"_32),("A"_13,"A"_23,"A"_33)]`
`1/-3 [(-3,0,0),(3,-1,0),(-9,-2,3)]`
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