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If A is a square matrix such that A2=AA^2=A and I is the identify matrix of the same order as A then (I+2A)3(I + 2A)^3-6A is equal to

Solution

✅ Correct Option: 3

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

Since A2=AA^2 = A, any occurrence of A2A^2 can be replaced with AA.


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

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

=I⋅I+I⋅2A+2A⋅I+2A⋅2A= I \cdot I + I \cdot 2A + 2A \cdot I + 2A \cdot 2A

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

=I+2A+2A+4A= I + 2A + 2A + 4A

=I+8A= I + 8A


Expand (I+2A)3(I + 2A)^3:

(I+2A)3=(I+2A)2×(I+2A)(I + 2A)^3 = (I + 2A)^2 \times (I + 2A)

=(I+8A)(I+2A)= (I + 8A)(I + 2A)

=I⋅I+I⋅2A+8A⋅I+8A⋅2A= I \cdot I + I \cdot 2A + 8A \cdot I + 8A \cdot 2A

=I+2A+8A+16A2= I + 2A + 8A + 16A^2

=I+2A+8A+16A= I + 2A + 8A + 16A

=I+26A= I + 26A


Calculate (I+2A)3−6A(I + 2A)^3 - 6A:

(I+2A)3−6A=(I+26A)−6A(I + 2A)^3 - 6A = (I + 26A) - 6A

=I+26A−6A= I + 26A - 6A

=I+20A= I + 20A

Therefore, (I+2A)3−6A=I+20A(I + 2A)^3 - 6A = I + 20A

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