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)] and I={:[(1,0),(0,1)] Show that, 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_(2).

Verify that A=[[a,b],[c,d]] satisfies the equation A^2-(a+d)A+(ad-bc) I=0 where I is the 2 x 2 unit matrix.

If a, b, c, d in R and A=[(a,b),(c,d)] , show that A^2-(a+d)A+(ad-bc)I is a null matrix.

If A=[[0,1],[1,0]],B=[[0,-i],[i,0]] and C=[[i,0],[0,-i]] , show that A^2=B^2=-C^2=I_2 and AB=-BA,AC=-CA and BC=-CB .

If A=[[0,1],[1,0]],B=[[0,-i],[i,0]] and C=[[i,0],[0,-i]] , show that A^2=B^2=-C^2=I_2 and AB=-BA,AC=-CA and BC=-CB .

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

Let A=[[a,bc,d]] and B=([p,q])!=[[0,0]] Given that AB,=B and a+d=2. Then the value of (ad-bc) is