Question
If \(A\) and \(B\) are square matrices of the same order, explain why in general \[ (A + B)(A – B) \ne A^2 – B^2. \]
Solution
Step 1: Expand
\[ (A + B)(A – B) \] \[ = A(A – B) + B(A – B) \] \[ = A^2 – AB + BA – B^2 \]Step 2: Compare with Algebra
In algebra: \[ (a+b)(a-b) = a^2 – b^2 \] But for matrices: \[ AB \ne BA \quad \text{(in general)} \]Step 3: Key Reason
\[ -AB + BA \ne 0 \] So, \[ (A+B)(A-B) \ne A^2 – B^2 \]Step 4: Special Case
If: \[ AB = BA \] then: \[ -AB + BA = 0 \] and identity becomes valid.Final Answer
\[
(A+B)(A-B) = A^2 – AB + BA – B^2
\]
\[
\ne A^2 – B^2 \quad \text{(in general)}
\]
\[
\text{Because } AB \ne BA
\]