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If A is a square matrix such that A2=AA^2 = A and I is the identity matrix of same order as A, then the value of (A−2I)2−(2A+I)2+11A(A-2I)^2 - (2A + I)^2 + 11A is:

Solution

✅ Correct Option: 3

Given A2=AA^2 = A and II is the identity matrix of the same order as AA.

Need to find (A−2I)2−(2A+I)2+11A(A - 2I)^2 - (2A + I)^2 + 11A


Expanding (A−2I)2(A - 2I)^2:

(A−2I)2=A2−2A⋅I−2I⋅A+(2I)2(A - 2I)^2 = A^2 - 2A \cdot I - 2I \cdot A + (2I)^2

=A2−2A−2A+4I= A^2 - 2A - 2A + 4I

=A2−4A+4I= A^2 - 4A + 4I

Using A2=AA^2 = A:

(A−2I)2=A−4A+4I(A - 2I)^2 = A - 4A + 4I

=−3A+4I= -3A + 4I


Expanding (2A+I)2(2A + I)^2:

(2A+I)2=(2A)2+2(2A)⋅I+I2(2A + I)^2 = (2A)^2 + 2(2A) \cdot I + I^2

=4A2+4A+I= 4A^2 + 4A + I

Using A2=AA^2 = A:

(2A+I)2=4A+4A+I(2A + I)^2 = 4A + 4A + I

=8A+I= 8A + I


Substituting into the expression:

(A−2I)2−(2A+I)2+11A=(−3A+4I)−(8A+I)+11A(A - 2I)^2 - (2A + I)^2 + 11A = (-3A + 4I) - (8A + I) + 11A

=−3A+4I−8A−I+11A= -3A + 4I - 8A - I + 11A

=(−3−8+11)A+(4−1)I= (-3 - 8 + 11)A + (4 - 1)I

=0A+3I= 0A + 3I

=3I= 3I

Therefore, the value is 3I3I.

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