If A is a square matrix such that A² = I, then (A – I)³ + (A + I)³ – 7A is equal to

The correct option is A.

To find the value of the expression, we first recall the properties of the identity matrix $I$ and how it interacts with any square matrix $A$ of the same order:
1. $AI=IA=A$ (the identity matrix commutes with all matrices)
2. $I^n=I$ for any positive integer $n$

Because the matrix $A$ and the identity matrix $I$ commute, we can expand powers of their sum and difference using the standard algebraic binomial expansion formulas.

Let us expand the term ${(A-I)}^3$:
$${(A-I)}^3=A^3-3A^{2}I+3A{I}^2-I^3$$
Using the properties of the identity matrix, this simplifies to:
$${(A-I)}^3=A^3-3A^2+3A-I$$

Next, let us expand the term ${(A+I)}^3$:
$${(A+I)}^3=A^3+3A^{2}I+3A{I}^2+I^3$$
Which simplifies to:
$${(A+I)}^3=A^3+3A^2+3A+I$$

Now, we add these two expanded equations together:
$${(A-I)}^3+{(A+I)}^3=(A^3-3A^2+3A-I)+(A^3+3A^2+3A+I)$$
Grouping the like terms together:
$${(A-I)}^3+{(A+I)}^3=2A^3+6A$$

We are given the condition that $A^2=I$. We can use this to simplify $A^3$:
$$A^3=A^2\cdot A=I\cdot A=A$$

Substituting $A^3=A$ back into our summed expression:
$${(A-I)}^3+{(A+I)}^3=2A+6A=8A$$

Finally, we subtract $7A$ from this result as specified in the original question:
$${(A-I)}^3+{(A+I)}^3-7A=8A-7A=A$$

Thus, the expression simplifies to $A$.