Question
If \(A\) and \(B\) are square matrices of the same order, explain why in general \[ (A – B)^2 \ne A^2 – 2AB + B^2. \]
Solution
Step 1: Expand \((A-B)^2\)
\[ (A – B)^2 = (A – B)(A – B) \] \[ = A^2 – AB – BA + B^2 \]Step 2: Compare with Algebra
In algebra: \[ (a-b)^2 = a^2 – 2ab + b^2 \] But for matrices: \[ AB \ne BA \quad \text{(in general)} \]Step 3: Key Reason
\[ -AB – BA \ne -2AB \] because matrix multiplication is **not commutative**.Step 4: Special Case
If: \[ AB = BA \] then: \[ -AB – BA = -2AB \] and the identity becomes valid.Final Answer
\[
(A-B)^2 = A^2 – AB – BA + B^2
\]
\[
\ne A^2 – 2AB + B^2 \quad \text{(in general)}
\]
\[
\text{Because } AB \ne BA
\]