# True or False: $(A-B)(A+B)=A^2-B^2$ for Matrices $A$ and $B$ ## Problem 96

Let $A$ and $B$ be $2\times 2$ matrices.

Prove or find a counterexample for the statement that $(A-B)(A+B)=A^2-B^2$. Add to solve later

## Hint.

In general, matrix multiplication is not commutative: $AB$ and $BA$ might be different.

## Solution.

Let us calculate $(A-B)(A+B)$ as follows using the fact that the matrix product is distributive.
\begin{align*}
(A-B)(A+B)&=A(A+B)-B(A+B)\\
&=A^2+AB-BA-B^2\\
&=A^2-B^2+(AB-BA).
\end{align*}
Thus if $(A-B)(A+B)=A^2-B^2$ then $AB-BA=O$, the zero matrix. Equivalently, $AB=BA$.

Note that matrix multiplication is not commutative, namely, $AB\neq BA$ in general.
Thus we can disprove the statement if we find matrices $A$ and $B$ such that $AB \neq BA$.

For example, let
$A=\begin{bmatrix} 1 & 1\\ 0& 0 \end{bmatrix} \text{ and } B=\begin{bmatrix} 0 & 1\\ 0& 1 \end{bmatrix}.$ Then we have
$AB=\begin{bmatrix} 0 & 2\\ 0& 0 \end{bmatrix} \text{ and } BA=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}.$

Since $AB \neq BA$, we have $(A-B)(A+B) \neq A^2-B^2$ for $A=\begin{bmatrix} 1 & 1\\ 0& 0 \end{bmatrix} \text{ and } B=\begin{bmatrix} 0 & 1\\ 0& 1 \end{bmatrix}.$

Hence we found a counterexample for the statement.

## 10 True or False Quiz Problems about Matrix Operations

Check out the post “10 True or False Problems about Basic Matrix Operations” and take a quiz about basic properties of matrix operations.

There are 10 True or False problems about basic properties of matrix operations (matrix product, transpose, etc.).

The complete solutions are given as well. Add to solve later

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1. 07/19/2017

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