Q.Using the same matrices $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}5&6\\7&8\end{pmatrix}$ from the previous example, find $BA$ and verify that $AB \neq BA$.

Computing the product in the reverse order, $BA$, and comparing it with the previously found $AB$ demonstrates that matrix multiplication does not commute.

Row 1 of $B$ with each column of $A$

$(1,1)$: $5(1)+6(3)=5+18=23$.

$(1,2)$: $5(2)+6(4)=10+24=34$.

Row 2 of $B$ with each column of $A$

$(2,1)$: $7(1)+8(3)=7+24=31$.

$(2,2)$: $7(2)+8(4)=14+32=46$.

Comparing with $AB$

From the previous worked example, $AB=\begin{pmatrix}19&22\\43&50\end{pmatrix}$. Comparing entry by entry, $23\neq19$, $34\neq22$, $31\neq43$, $46\neq50$ — every single entry differs, so $AB\neq BA$, confirming matrix multiplication is NOT commutative in general. …

Assuming $BA$ must simply be $AB$ 'flipped' or 'reversed' in some simple pattern (like transposing it) — there is no such shortcut; $BA$ must be computed independently using its o …