Question

Let $A$ be a matrix such that $A^2 = I$, where $I$ is an identity matrix. Then $(I + A)^4 - 8A$ is equal to:

• A. $8 I$
• B. $5 I$
• C. 8(I + A)
• D. 5(I - A)

Answer 1

Answer: (A)

$8 I$

Answer 2

Given $A^2 = I$, the identity matrix, we analyze $(I + A)^4$:

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

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

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

$(2I + 2A)^2 = 4I^2 + 8IA + 4A^2$

Since $A^2 = I$, substitute $4A^2 = 4I$:

$(2I + 2A)^2 = 4I + 8A + 4I = 8I + 8A$

Now compute $(I + A)^4 - 8A$:

$(I + A)^4 - 8A = (8I + 8A) - 8A = 8I$

Thus, the result is: $8I$