1. **Problem Statement:** We want to check if the equality $$(A + B)(A - B) = A^2 - B^2$$ holds for general $n \times n$ matrices $A$ and $B$. 2. **Recall the formula for difference of squares:** For real numbers, we have $$(x + y)(x - y) = x^2 - y^2.$$ This relies on commutativity of multiplication. 3. **Important note:** Matrix multiplication is generally **not commutative**, meaning $$AB \neq BA$$ in general. 4. **Expand the left side:** $$ (A + B)(A - B) = A A - A B + B A - B B = A^2 - AB + BA - B^2 $$ 5. **Compare with the right side:** $$ A^2 - B^2 $$ 6. **Conclusion:** The equality holds if and only if $$AB = BA$$ (i.e., $A$ and $B$ commute). 7. **Example to show it is not true in general:** Let $$ A = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}, \quad B = \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} $$ Calculate: $$ AB = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} $$ $$ BA = \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} $$ Since $$AB \neq BA$$, the equality does not hold. 8. **Verify the original expression:** $$ (A + B)(A - B) = A^2 - AB + BA - B^2 $$ Calculate each term: $$ A^2 = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}^2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} $$ $$ B^2 = \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}^2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} $$ So, $$ A^2 - B^2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} $$ Calculate left side: $$ A^2 - AB + BA - B^2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} - \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} + \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} - \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = \begin{bmatrix}-1 & 0 \\ 0 & 1\end{bmatrix} $$ This is not equal to $$A^2 - B^2$$. **Final answer:** The equality $$(A + B)(A - B) = A^2 - B^2$$ is **not true in general** for matrices unless $A$ and $B$ commute.