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Question
Find the inverse of the following matrix (if they exist).
`[(1,3,-2),(-3,0,-5),(2,5,0)]`
Solution
Let A = `[(1,3,-2),(-3,0,-5),(2,5,0)]`
∴ |A| = `|(1,3,-2),(-3,0,-5),(2,5,0)|`
= 1(0 + 25) + 3(0 + 10) + 2(- 15 - 0)
= 25 + 30 - 30
= 25 ≠ 0
∴ A-1 exists.
Consider AA-1 = I
∴ `[(1,3,-2),(-3,0,-5),(2,5,0)] "A"^-1 = [(1,0,0),(0,1,0),(0,0,1)]`
By R2 → R2 + 3R1
`[(1,3,-2),(0,9,-11),(2,5,0)] "A"^-1 = [(1,0,0),(3,1,0),(0,0,1)]`
By R3 → R3 - 2R1
`[(1,3,-2),(0,9,-11),(0,-1,4)] "A"^-1 = [(1,0,0),(3,1,0),(-2,0,1)]`
By `"R" -> 1/9 "R"_2`
`[(1,3,-2),(0,1,-11/9),(0,-1,4)] "A"^-1 = [(1,0,0),(1/3,1/9,0),(-2,0,1)]`
By `"R"_3 -> "R"_3 + "R"_2`
`[(1,3,-2),(0,1,-11/9),(0,0,25/9)] "A"^-1 = [(1,0,0),(1/3,1/9,0),(-5/3,1/9,1)]`
By `"R"_1 -> "R"_1 + 3"R"_2`
`[(1,0,5/3),(0,1,-11/9),(0,0,25/9)] "A"^-1 = [(0,-1/3,0),(1/3,1/9,0),(-5/3,1/9,1)]`
By `"R"_3 -> 9/25 "R"_3`
`[(1,0,5/3),(0,1,-11/9),(0,0,1)] "A"^-1 = [(0,-1/3,0),(1/3,1/9,0),(-3/5,1/25,9/25)]`
By `"R"_2 -> "R"_2 + 11/9 "R"_3`
`[(1,0,5/3),(0,1,0),(0,0,1)] "A"^-1 = [(0,-1/3,0),(-2/5,4/25,11/25),(-3/5,1/25,9/25)]`
By `"R"_1 -> "R"_1 - 5/3 "R"_3`
`[(1,0,0),(0,1,0),(0,0,1)] "A"^-1 = [(5/5,-6/15,-3/5),(-2/5,4/25,11/25),(-3/5,1/25,9/25)]`
`"A"^-1 = [(1,-2/5,-3/5),(-2/5,4/25,11/25),(-3/5,1/25,9/25)]`
`"A"^-1 = 1/25 [(25,-10,-15),(-10,4,11),(-15,1,9)]`
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