Q.If $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}2&0\\1&2\end{pmatrix}$, find $AB$ and $BA$, and verify that $AB \ne BA$.

Both products are defined since $A$ and $B$ are both $2\times2$, but matrix multiplication is not commutative in general, so $AB$ and $BA$ must be computed separately and compared.

Computing $AB$: using row-of-$A$ into column-of-$B$,

Row 1: $(1)(2)+(2)(1)=2+2=4$ and $(1)(0)+(2)(2)=0+4=4$.

Row 2: $(3)(2)+(4)(1)=6+4=10$ and $(3)(0)+(4)(2)=0+8=8$.

So $AB=\begin{pmatrix}4&4\\10&8\end{pmatrix}$.

Computing $BA$: using row-of-$B$ into column-of-$A$,

Row 1: $(2)(1)+(0)(3)=2+0=2$ and $(2)(2)+(0)(4)=4+0=4$.

Row 2: $(1)(1)+(2)(3)=1+6=7$ and $(1)(2)+(2)(4)=2+8=10$.

So $BA=\begin{pmatrix}2&4\\7&10\end{pmatrix}$.

Comparing entry by entry, $AB$'s $(1,1)$ entry is $4$ while $BA$'s is $2$ — already different, so $AB \ne BA$ without needing to compare every entry.

Independent check: recompute the $(2,1)$ entry of $AB$ using the general $2\times2$ formula from Section 3, $(AB)_{21}=ce+dg$ with $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}e&f\\g&h\end{pmatrix}=\begin{pmatrix}2&0\\1&2\end{pmatrix}$: $ce+dg=(3)(2)+(4)(1)=6+4=10$, matching the value found above.

Multiplying entries in the same position (like ordinary elementwise multiplication) instead of the row-into-column rule; also assuming $AB=BA$ without actually checking, which is wrong for matrices in general.