Question

Let \( A = \begin{bmatrix} 1 & 2 2 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} x & y 1 & 2 \end{bmatrix} \) be two matrices such that \( (A + B)(A - B) = A^2 - B^2 \).
If \( C = \begin{bmatrix} x & 2 1 & y \end{bmatrix} \), then trace \((C) = \)

• A. \( 3 \)
• B. \( 5 \)
• C. \( 7 \)
• D. \( 9 \)

Hint:

Whenever you see the identity \( (A+B)(A-B) = A^2 - B^2 \), immediately jump to the condition \( AB = BA \). This is one of the most common ways matrix commutativity is tested in entrance exams.

Answer 1

The Correct Option is A

Concept: For any two square matrices \( A \) and \( B \), the expansion of \( (A + B)(A - B) \) is given by \( A^2 - AB + BA - B^2 \). The identity \( (A + B)(A - B) = A^2 - B^2 \) holds true if and only if \( AB = BA \), meaning the matrices commute. The trace of a matrix is the sum of the elements on its main diagonal.

Step 1: Using the commutativity condition \( AB = BA \).
Expanding \( (A + B)(A - B) \): \[ (A + B)(A - B) = A^2 - AB + BA - B^2 \] Given \( (A + B)(A - B) = A^2 - B^2 \), it implies: \[ -AB + BA = 0 \quad \Rightarrow \quad AB = BA \]

Step 2: Calculating products \( AB \) and \( BA \).
\[ AB = \begin{bmatrix} 1 & 2 2 & 1 \end{bmatrix} \begin{bmatrix} x & y 1 & 2 \end{bmatrix} = \begin{bmatrix} x + 2 & y + 4 2x + 1 & 2y + 2 \end{bmatrix} \] \[ BA = \begin{bmatrix} x & y 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 2 & 1 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y 1 + 4 & 2 + 2 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y 5 & 4 \end{bmatrix} \]

Step 3: Equating \( AB = BA \) to find \( x \) and \( y \).
Comparing the elements: 1) \( x + 2 = x + 2y \Rightarrow 2y = 2 \Rightarrow y = 1 \) 2) \( 2y + 2 = 4 \Rightarrow 2(1) + 2 = 4 \) (Consistent) 3) \( 2x + 1 = 5 \Rightarrow 2x = 4 \Rightarrow x = 2 \)

Step 4: Finding the trace of matrix \( C \).
Matrix \( C = \begin{bmatrix} x & 2 1 & y \end{bmatrix} = \begin{bmatrix} 2 & 2 1 & 1 \end{bmatrix} \). The trace of \( C \) is the sum of the diagonal elements: \[ \text{trace}(C) = x + y = 2 + 1 = 3 \]