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Byju's Answer
Standard XII
Mathematics
Finding Inverse Using Elementary Transformations
Use the metho...
Question

Use the method of elementary row transformation to compute the inverse of
⎡⎢⎣125231−111⎤⎥⎦

A
A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣22117−1321−172737521−17−121⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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B
A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣12117−1121−172737521−27−221⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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C
A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣42117−1621−172737521−27−421⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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D
A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣42127−1321−172737421−27−121⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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Solution

The correct option is A A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣22117−1321−172737521−17−121⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
Let A=⎡⎢⎣125231−111⎤⎥⎦
⇒Write AA−1=I
⎡⎢⎣125231−111⎤⎥⎦A−1=⎡⎢⎣100010001⎤⎥⎦
R21(−2)~R31(1)⎡⎢⎣105231−111⎤⎥⎦A−1=⎡⎢⎣100−210101⎤⎥⎦
R2(−1)~R3(1/3)⎡⎢⎣125019012⎤⎥⎦A−1=⎡⎢ ⎢ ⎢⎣1002−1013013⎤⎥ ⎥ ⎥⎦
R12(−2)~R32(−1)⎡⎢⎣10−1301900−7⎤⎥⎦A−1=⎡⎢ ⎢ ⎢⎣−3202−10−53113⎤⎥ ⎥ ⎥⎦
R3(−1/7)~⎡⎢⎣10−13019001⎤⎥⎦A−1=⎡⎢ ⎢ ⎢⎣−3202−10521−17−121⎤⎥ ⎥ ⎥⎦
R13(13)~R23(−9)⎡⎢⎣100010001⎤⎥⎦A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣22117−1321−172737521−17−121⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
Hence, A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣22117−1321−172737521−17−121⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

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