A=([a,b],[c,d])_(" then "AdjA=)

If A = [[a,b],[c,d]] then adj(adjA) is equal to; i] adjA ii] A iii] A^T iv] -A

If A=[[a, b],[ c ,d]] , then a d j\ A is [[-d,-b],[-c, a]] (b) [[d,-b],[-c ,a]] (c) [[d, b],[ c, a]] (d) [[d, c],[ b ,a]]

If A=[[a, b],[ c ,d]] , then a d j\ A is (a) [[-d,-b],[-c, a]] (b) [[d,-b],[-c ,a]] (c) [[d, b],[ c, a]] (d) [[d, c],[ b ,a]]

If A=[(a,b),(c,d)] , then adj(adjA) is equal to

(a)If A=[a,b],B=[c,d],C=[d,e] then {(a,c),(a,d),(a,e),(b,c),(b,d),(b,e)}=………

If A=[(2,0,0),(0,cos,sinx),(0,-sinx,cosx)] then (AdjA)^-1= (A) 1/2A (B) A (C) 2A (D) 4A

If A=[(2,0,0),(0,cos,sinx),(0,-sinx,cosx)] then (AdjA)^-1= (A) 1/2A (B) A (C) 2A (D) 4A

If A=[(a;b);(c;d)]; find adjA

If A = [(a,b),(c,d)] show that, (i) adj (adj.A) = A (ii). (A^(-1))^(-1) = A .