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Question
Find the inverse of the following matrix (if they exist):
`[(2,-3,3),(2,2,3),(3,-2,2)]`
Solution
Let A = `[(2,−3,3),(2,2,3),(3,−2,2)]`
∴ |A| = `|(2,-3,3),(2,2,3),(3,−2,2)|`
= 2(4 + 6) + 3(4 − 9) + 3(− 4 − 6)
= 20 − 15 − 30
= − 25 ≠ 0
∴ A-1 exists.
Consider AA-1 = I
∴ `[(2,-3,3),(2,2,3),(3,-2,2)] "A"^-1 = [(1,0,0),(0,1,0),(0,0,1)]`
By R1↔R3, we get,
`[(3,-2,2),(2,2,3),(2,-3,3)] "A"^-1 = [(0,0,1),(0,1,0),(1,0,0)]`
By R1 → R1 − R2, we get
`[(1,-4,-1),(2,2,3),(2,-3,3)] "A"^-1 = [(0,-1,1),(0,1,0),(1,0,0)]`
By R2 → R2 − 2R1 and R3 → R3 − 2R1, we get,
`[(1,-4,-1),(0,10,5),(0,5,5)] "A"^-1 = [(0,-1,1),(0,3,-2),(1,2,-2)]`
By R2 → `(1/10) "R"_2`, we get,
`[(1,-4,-1),(0,1,1/2),(0,5,5)] "A"^-1 = [(0,-1,1),(0,3/10,-1/5),(1,2,-2)]`
By R1 → R1 + 4R2 and R3 → R3 - 5R2, we get,
`[(1,0,1),(0,1,1/2),(0,0,5/2)] "A"^-1 = [(0,1/5,1/5),(0,3/10,-1/5),(1,1/2,-1)]`
By R3 → `(2/5)"R"_3`, we get,
`[(1,0,1),(0,1,1/2),(0,0,1)] "A"^-1 = [(0,1/5,1/5),(0,3/10,-1/5),(2/5,1/5,-2/5)]`
By R1 →R1 − R3 and R2 → `"R"_2 - 1/2"R"_3`, we get,
`[(1,0,0),(0,1,0),(0,0,1)] "A"^-1 = [(-2/5,0,3/5),(-1/5,1/5,0),(2/5,1/5,-2/5)]`
∴ `"A"^-1 = [(-2/5,0,3/5),(-1/5,1/5,0),(2/5,1/5,-2/5)]`
Notes
Answer in the textbook is incorrect.
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