Question

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

• A. A
• B. 2A
• C. 3A
• D. 5A

Hint:

When dealing with matrix polynomials involving the identity matrix I, remember that I behaves like the number 1 in scalar algebra. It commutes with any matrix, and \(I^n = I\). This allows the use of standard algebraic formulas like the binomial theorem.

Answer 1

The Correct Option is D

Step 1: Conceptual Foundation:

This problem requires the simplification of a matrix polynomial. The method leverages properties of matrix algebra, specifically the given condition \( A^2 = I \). Since matrices \( A \) and the identity matrix \( I \) commute (i.e., \( AI = IA = A \)), standard binomial expansion principles are applicable.

Step 2: Core Formulas/Methodology:

The binomial expansion formulas utilized are:

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

Step 3: Detailed Derivation:

Expansion of the terms \( (A - I)^3 \) and \( (A + I)^3 \) is performed:

\[ (A - I)^3 = A^3 - 3A^2I + 3AI^2 - I^3 \]

Applying \( I^n = I \) and \( A^2 = I \), this simplifies to:

\[ (A - I)^3 = A^3 - 3A^2 + 3A - I \]

The second term expands as:

\[ (A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3 = A^3 + 3A^2 + 3A + I \]

Summing these two expansions yields:

\[ (A - I)^3 + (A + I)^3 = (A^3 - 3A^2 + 3A - I) + (A^3 + 3A^2 + 3A + I) \]

\[ = 2A^3 + 6A \]

Given \( A^2 = I \), we derive \( A^3 \):

\[ A^3 = A^2 \cdot A = I \cdot A = A \]

Substitution of \( A^3 = A \) into the summed expression gives:

\[ (A - I)^3 + (A + I)^3 = 2(A) + 6A = 8A \]

The final operation involves subtracting the specified term:

\[ (A - I)^3 + (A + I)^3 - 3A = 8A - 3A = 5A \]

Step 4: Conclusion:

The simplified value of the expression is 5A.