If A is a square matrix and I is the identity matrix of same order such that A² = I, then (A - I)³ + (A + I)³ - 3A is equal to

The correct option is 5A.

Let's verify this step-by-step using matrix algebra. We are given that $A^2=I$, where $A$ is a square matrix and $I$ is the identity matrix of the same order. Since matrix multiplication with the identity matrix commutes (meaning $AI=IA=A$), we can expand the algebraic expressions ${(A-I)}^3$ and ${(A+I)}^3$ using the binomial expansion just like ordinary algebraic variables.

First, let's expand ${(A-I)}^3$:
$${(A-I)}^3=A^3-3A^{2}I+3A{I}^2-I^3$$
Since $I^2=I$, $I^3=I$, and $AI=A$, we simplify this to:
$${(A-I)}^3=A^3-3A^2+3A-I$$

Next, let's expand ${(A+I)}^3$:
$${(A+I)}^3=A^3+3A^{2}I+3A{I}^2+I^3$$
Simplifying using the properties of the identity matrix yields:
$${(A+I)}^{\mathrm{3<!--="true"-->}}=A^3+3A^2+\mathrm{3<!--="true"--><mi>A</mi><mo>+</mo><mi>I</mi>}$$

Now, let's add these two expanded expressions together:
$${(A-I)}^3+{(A+I)}^3=(A^3-3A^2+3A-I)+(A^3+3A^2+3A+I)$$
Combining the like terms:
$${(A-I)}^3+{(A+I)}^3=2A^3+6A$$

We are given that $A^2=I$. Therefore, we can simplify $A^3$ as follows:
$$A^3=A^2·A=I·A=A$$
Substituting $A^3=A$ back into our expression:
$${(A-I)}^3+{(A+I)}^3=2A+6A=8A$$

Finally, we subtract $3A$ from the combined sum as required by the problem statement:
$${(A-I)}^3+{(A+I)}^3-3A=8A-3A=5A$$

Thus, the final simplified expression is equal to $5A$.