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 If 2    13    2A3253=1    00    1, then the matrix A=

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a
1110
b
1101
c
1011
d
0111

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detailed solution

Correct option is A

2132A−325−3=1001⇒A=2132−11001−325−3−1=2−1−3210011(9−10)−3−2−5−3=2−1−323253=6−54−3−9+10−6+6=1110


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