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Byju's Answer
Standard XII
Mathematics
Finding Inverse Using Elementary Transformations
The inverse o...
Question

The inverse of A=⎡⎢⎣30220−2011⎤⎥⎦ is

A
⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣15150−15310115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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B
⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−15151−15310115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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C
⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣2535−1−1535115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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D
⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−15250−−35310115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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Solution

The correct option is A ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣15150−15310115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

⎡⎢⎣30220−2011⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦⋅A

R1→R1+R2
⎡⎢⎣50020−2011⎤⎥⎦=⎡⎢⎣110010001⎤⎥⎦⋅A

R1→R15
⎡⎢⎣10020−2011⎤⎥⎦=⎡⎢ ⎢ ⎢⎣15150010001⎤⎥ ⎥ ⎥⎦⋅A

R2→R2−2R1
⎡⎢⎣10000−2011⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣15150−25350001⎤⎥ ⎥ ⎥ ⎥ ⎥⎦⋅A

R2→−12R2
⎡⎢⎣100001011⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣1515015−3100001⎤⎥ ⎥ ⎥ ⎥ ⎥⎦⋅A

R2↔R3
⎡⎢⎣100011001⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣1515000115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥⎦⋅A

R2→R2−R3
⎡⎢⎣100010001⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣15150−15310115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦⋅A
This is of the form I=A−1⋅A
Thus A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣15150−15310115−3100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦


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