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Question
Using elementary row transformation, find the inverse of the matrix
`[(2,-3,5),(3,2,-4),(1,1,-2)]`
Solution
We have A =`[(2,-3,5),(3,2,-4),(1,1,-2)]`
A=IA `[(2,-3,5),(3,2,-4),(1,1,-2)]=[(1,0,0),(0,1,0),(1,0,0)]`A
R1 ↔ R3
`[(1,1,-2),(3,2,-4),(2,-3,5)]=[(0,0,1),(0,1,0),(1,0,0)]`A
R2 → R2 - 3R1,
R3 → R3 - 2R1
`[(1,1,-2),(0,-1,2),(0,-5,9)]=[(0,0,1),(0,1,-3),(1,0,-2)]`A
R2 → R1 +R2 ,
R3 → R3 - 5R2
`[(1,0,0),(0,-1,2),(0,0,-1)]=[(0,0,-2),(0,1,-3),(1,-5,13)]`A
R2 → - R2 ,
R3 → - R3
`[(1,0,0),(0,1,-2),(0,0,1)]=[(0,1,-2),(0,-1,3),(-1,5,-13)]`A
R2 → R2 + 2R3
`[(1,0,0),(0,1,0),(0,0,1)]=[(0,1,-2),(-2,9,-23),(-1,5,-13)]`A
We know, I = AA-1
Therefore, inverse of A i.e. `"A"^-1 = [(0,1,-2),(-2,9,-23),(-1,5,-13)]`
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