[" (1) If "A=[[a,b],[c,d]],quad I=[[1,0],[0,1]]" then "A^(2)-(a+d)A=],[" 1) "O," 2) "I," -3) "(bc-ad)I]

If A= [[a,b],[c,d]] and I= [[1,0],[0,1]] then A^(2)-(a+d)A=

If A=[[a,b],[c,d]] and I=[[1,0],[0,1]] then A^2-(a+d)A-(bc-ad)I=

If A=[(a,b),(c,d)],I=[(1,0),(0,1)] then A^(2)-(a+d)A=

If A={:[(a,b),(c,d)] and I={:[(1,0),(0,1)] Show that, A^(2)-(a+d)A=(bc-ad)I .

If A=((a, b), (c, d)) and I=((1, 0), (0, 1)) show that A^(2)-(a+d)A=(bc-ad)I_(2).

If A=[(5,2),(-1,2)] and I=[(1,0),(0,1)] show that : (A-3I)(A-4I)=O

If A=[{:(4,1),(3,2):}] and I=[{:(1,0),(0,1):}] then A^(2)-6A is equal to : a) 3I b) -5I c) 5I d) none of these

If A= [(1,0),(1,1)] then A^(n)+nI is equal to a) I b)nA c)nA-I d)I+nA

If A = ((a,b),(c,d)) , such that ad - bc != 0 then A^(-1) is :

If A=[(1,-1,0),(1,0,0),(0,0,-1)] , then A^(-1) is...a) A^T b) A^2 c) A d) I