This problem involves simplifying a matrix expression using the given property that for a square matrix A and the identity matrix I of the same order, A² = I.

Matrix Algebra Simplification Using A² = I

We are asked to simplify the expression: $ (A - I)^3 + (A + I)^3 - 3A $ We know the binomial expansion formulas:

• $ (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 $
• $ (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 $

Let's apply these to the matrix expression, substituting x = A and y = I.

Expanding the Cubic Terms

First, consider the term $(A - I)^3$. Applying the formula:

$ (A - I)^3 = A^3 - 3A^2I + 3AI^2 - I^3 $

Now, we use the properties of matrices and the identity matrix:

• $ A^2 = I $ (Given)
• $ A^3 = A^2 \cdot A = I \cdot A = A $
• $ AI = IA = A $
• $ I^2 = I \cdot I = I $
• $ I^3 = I^2 \cdot I = I \cdot I = I $

Substitute these properties into the expansion:

$ (A - I)^3 = A - 3(I)I + 3A(I) - I $ $ (A - I)^3 = A - 3I + 3A - I $ $ (A - I)^3 = (A + 3A) + (-3I - I) $ $ (A - I)^3 = 4A - 4I $

Next, consider the term $(A + I)^3$. Applying the formula:

$ (A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3 $

Using the same properties as above:

$ (A + I)^3 = A + 3(I)I + 3A(I) + I $ $ (A + I)^3 = A + 3I + 3A + I $ $ (A + I)^3 = (A + 3A) + (3I + I) $ $ (A + I)^3 = 4A + 4I $

Simplifying the Full Expression

Now substitute the simplified cubic terms back into the original expression:

$ (A - I)^3 + (A + I)^3 - 3A = (4A - 4I) + (4A + 4I) - 3A $

Combine the terms:

$ = 4A - 4I + 4A + 4I - 3A $

Group the A terms and the I terms:

$ = (4A + 4A - 3A) + (-4I + 4I) $ $ = (8A - 3A) + (0) $ $ = 5A $

Therefore, the expression $(A - I)^3 + (A + I)^3 - 3A$ simplifies to 5A.