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Question
Find the inverse of the following matrix, using elementary transformations:
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
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Solution
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
AA-1 =I
`[[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]] A^(-1) = [[1 , 0 , 0 ],[0 , 1 , 0],[0 , 0 ,1]]`
R2 → R2 - R1
R3 → R3 - R1
`[[2 , 3 , 1 ],[0 , 1 , 0],[1, 4 ,1]] A^(-1) = [[1 , 0 , 0 ],[-1 , 1 , 0],[-1 , 0 ,1]]`
R1 ↔ R3
`[[1 , 4 , 1 ],[0 , 1 , 0],[2 , 3 ,1]] A^(-1) = [[-1 , 0 , 1 ],[-1 , 1 , 0],[1 , 0 ,0]]`
R3 → R3 - 2R1
`[[1 , 4 , 1 ],[0 , 1 , 0],[0 , -5 ,-1]] A^(-1) = [[-1 , 0 , 1 ],[-1 , 1 , 0],[3 , 0 ,-2]]`
R1 → R1 - 4R2
R3 → R3 - 5R2
`[[1 , 0 , 1 ],[0 , 1 , 0],[0 , 0 ,-1]] A^(-1) = [[3 , -4 , 1 ],[-1 , 1 , 0],[-2 , 5 ,-2]]`
`R_1 ->R_1 +R_3`
`[[1 , 0 , 0 ],[0 , 1 , 0],[0 , 0 ,-1]] A^(-1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[-2 , 5 ,-2]]`
`R_3 -> -R_3`
`[[1 , 0 , 0 ],[0 , 1 , 0],[0 , 0 ,1]] A^(-1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[2 , -5 ,2]]`
`I .A^(1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[2 , -5 ,2]]`
` ⇒A^(1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[2 , -5 ,2]]`
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