If A is a square matrix and $I$ is an identity matrix of same order such that $A^2 = A$, then $(I+A)^3 - 8I$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $7(A-I)$
Given: $A^{2} = A$ and $I$ is the identity matrix of the same order.
We need to find $(I + A)^{3} - 8I$.
Expand using the binomial theorem:
$(I + A)^{3} = I^{3} + 3I^{2}A + 3IA^{2} + A^{3}$
Since $I^{2} = I$ and $IA = A$, this simplifies to:
$(I + A)^{3} = I + 3A + 3A^{2} + A^{3}$
Given $A^{2} = A \;\Rightarrow\; A^{3} = A^{2}A = A \cdot A = A$.
Substitute these values:
$(I + A)^{3} = I + 3A + 3A + A = I + 7A$
Now, $(I + A)^{3} - 8I = (I + 7A) - 8I = 7A - 7I = 7(A - I)$
Final Answer: $(I + A)^{3} - 8I = 7(A - I)$