Given A2=A and I is the identity matrix of the same order as A.
Need to find (A−2I)2−(2A+I)2+11A
Expanding (A−2I)2:
(A−2I)2=A2−2A⋅I−2I⋅A+(2I)2
=A2−2A−2A+4I
=A2−4A+4I
Using A2=A:
(A−2I)2=A−4A+4I
=−3A+4I
Expanding (2A+I)2:
(2A+I)2=(2A)2+2(2A)⋅I+I2
=4A2+4A+I
Using A2=A:
(2A+I)2=4A+4A+I
=8A+I
Substituting into the expression:
(A−2I)2−(2A+I)2+11A=(−3A+4I)−(8A+I)+11A
=−3A+4I−8A−I+11A
=(−3−8+11)A+(4−1)I
=0A+3I
=3I
Therefore, the value is 3I.