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B
[51638472]
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C
[51846372]
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D
[72−63−8451]
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Solution
The correct option is B[51638472] We have, A=[2−3−41]∴A2=A.A[2−3−41][2−3−41]=[4+12−6−3−8−412+1]=[16−9−1213]Now,3A2+12A=3[16−9−1213]+12[2−3−41]=[48−27−3639]+[24−364812]=[72−63−8451]∴adj(3A2+12A)=[51638472]