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

Computing the product in both orders and comparing shows that matrix multiplication does not commute.

Computing $AB$

$(1,1)$: $2(1)+1(2)=2+2=4$. $(1,2)$: $2(4)+1(1)=8+1=9$. $(2,1)$: $0(1)+3(2)=0+6=6$. $(2,2)$: $0(4)+3(1)=0+3=3$.

Computing $BA$

$(1,1)$: $1(2)+4(0)=2+0=2$. $(1,2)$: $1(1)+4(3)=1+12=13$. $(2,1)$: $2(2)+1(0)=4+0=4$. $(2,2)$: $2(1)+1(3)=2+3=5$.

Comparing $AB$ and $BA$

Entry by entry, $4\neq2$, $9\neq13$, $6\neq4$, $3\neq5$ — every single entry differs, so $AB\neq BA$, confirming matrix multiplication is NOT commutative in general. …

Assuming $BA$ can be obtained from $AB$ by some simple shortcut (like transposing it) — there is no such shortcut; $BA$ must be co …