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

The Correct Option is A

Given the matrix equation $A^2 = I$, it indicates that $A$ is an involutory matrix such that its own square is the identity matrix $I$. We are tasked with finding $(I + A)^4 - 8A$. Let us process this calculation step-by-step: 

First, let's simplify $(I + A)^2$. Expand the square:

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

This simplification utilizes $A^2 = I$. Now, compute $(I + A)^4$ using the result of $(I + A)^2$:

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

Further expand $(2I + 2A)^2$:

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

Now substitute $(I + A)^4 = 8I + 8A$ into the original expression $(I + A)^4 - 8A$:

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

Thus, the expression $(I + A)^4 - 8A$ simplifies to $8I$.

The correct answer is $8I$.

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:  \[8I\]