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Question

Using elementary operations, find the inverse of the following matrix: 112123311

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Solution

Let A=112123311
For applying elementary row operation we write,
A = I. A
112123311=100010001A
Applying R1R2, we get
123112311=010100001A
Applying R2R2+R1 and R3R33R1, we get
123035058=010110031A
Applying R1R123R2, we get
⎢ ⎢1013035058⎥ ⎥=⎢ ⎢23130110031⎥ ⎥A
Applying R213R2, we get
⎢ ⎢ ⎢10130153058⎥ ⎥ ⎥=⎢ ⎢ ⎢2313013130031⎥ ⎥ ⎥A
Applying R3R3+5R2, we get
⎢ ⎢ ⎢101301530013⎥ ⎥ ⎥=⎢ ⎢ ⎢ ⎢231301313053431⎥ ⎥ ⎥ ⎥A
Applying R1R1+R3 and R25R3
⎢ ⎢1000100013⎥ ⎥=⎢ ⎢11187553431⎥ ⎥A
Applying R33R3, we get
100010001=111875543A
Hence A1=111875543

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