Let M=01a1233b1 and adj M=-11-18-62-53-1 where a and b are real numbers.  Which of the following options is/are correct?

AdjM=2−3b8b-6ab-1-3a33-2aa-1T=2−3bab-13-2a8-3aab-63-1Comparing b−6=−5⇒b=1 it gives 3−2a=−1⇒a=2a+b=3Option 1 is correct. 'M=8+21−6=−2β=−1,γ=1adj   M2=M22=M4=16     Option 2 is correct adj   M−1+adj   M−1=2.MM=−M , option 3 is correct   since adj A-1=Adet A Mαβγ=β+2γα+2β+3γ3α+β+γ=123  so α+β+γ=1⇒2α=2⇒α=1 it gives β=−1,γ=1so α−β+γ=3 option 4 is correct