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Question

Using elementary transformation, find the inverse of the matrix 233223322.

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Solution

233223322A1=100010001
R1R3
322223233A1=001010100
R1R2
141223233A1=011010100
R22R1, R32R1
1410105055A1=011032122
R2/10
⎢ ⎢ ⎢1410112055⎥ ⎥ ⎥A1=01103/102/10122
R1+4R2,R35R2
⎢ ⎢ ⎢ ⎢ ⎢10101120052⎥ ⎥ ⎥ ⎥ ⎥A1=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢0151503102101121⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
R3×25,R1R3,R113R3,
100010001A1=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢2503515150251525⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
A1=15203110212.

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