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If AA is a square matrix such that A2=AA^2 = A and II is the identity matrix of the same order as AA, then (I+2A)2−5A(I + 2A)^2 - 5A is equal to

Solution

✅ Correct Option: 2

Given A2=AA^2 = A (idempotent matrix), we need to simplify (I+2A)2−5A(I + 2A)^2 - 5A.


Expanding (I+2A)2(I + 2A)^2 using the identity (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2:

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

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


Since A2=AA^2 = A:

=I+4A+4A= I + 4A + 4A

=I+8A= I + 8A


(I+2A)2−5A=I+8A−5A(I + 2A)^2 - 5A = I + 8A - 5A

=I+3A= I + 3A

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