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a
72−84−6351
b
51638472
c
51846372
d
72−63−8451
answer is B.
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Detailed Solution
We have __A=2−3−41∴A2=A⋅A=2−3−412−3−41=4+12−6−3−8−412+1=16−9−1213 Now, 3A2+12A=316−9−1213+122−3−41=48−27−3639+24−36−4812=72−63−8451∴ adj A2+12A=51638472