Q.Let $A$ and $B$ be square matrices of the order $3 \times 3$. Is $(AB)^2 = A^2 B^2$? Give reasons.

No. Since $(AB)^2 = ABAB$ but $A^2B^2 = AABB$, the two are equal only when $BA = AB$, and matrix multiplication is not commutative in general. A $3\times3$ counterexample: with

$A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},\quad B = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},$ …

The reasoning

By definition, $(AB)^2 = (AB)(AB) = ABAB$, while $A^2B^2 = (AA)(BB) = AABB$. These two agree only if the middle factors can be swapped, i.e. only if $BA = AB$. Since matrix multiplication is generally not commutative, the equality fails in general.

A concrete $3\times3$ counterexample

Take

First,

Next,

so

$A^2B^2 = \begin{bmatrix} 5 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$ …

Method: Testing whether a matrix "identity" is true — expand, then check commutativity

Many questions ask whether a familiar scalar identity (here $(AB)^2=A^2B^2$) carries over to matrices. The reliable method is to expand both sides by the definition of a power and see whether the orderings can only match if the matrices commute — then confirm with a counterexample.

Steps

Step 1: Expand each power as repeated multiplication, keeping the order.

Matrix multiplication is associative, so brackets can be dropped, but the left-to-right order must be preserved.

Step 2: Compare the two strings of factors.

$ABAB$ versus $AABB$ differ only in the middle: they are equal for all $A,B$ only if $BA=AB$, i.e. only if $A$ and $B$ commute. …

Common Mistakes

Mistake 1: Treating matrices like numbers and writing $(AB)^2=A^2B^2$ as automatically true.

Why it's wrong: the scalar rule relies on $ab=ba$; matrices generally satisfy $AB\ne BA$, so $ABAB$ cannot be rearranged into $AABB$. Correct approach: expand keeping order and note the identity needs commutativity.

Mistake 2: Picking a counterexample that accidentally gives equality and stopping there. …