Q.If $A=\begin{bmatrix}1 & 2 & 3\\2 & 4 & 6\\1 & 2 & 3\end{bmatrix}$, $B=\begin{bmatrix}1 & -1 & 1\\-3 & 2 & -1\\-2 & 1 & 0\end{bmatrix}$, show that $AB$ and $BA$ are both singular matrices.

Two matrices $A_{m\times n}$ and $B_{n\times p}$ can be multiplied — "are conformable for the product $AB$" — exactly when the number of columns of $A$ matches the number of rows of $B$; the product $AB$ then has order $m\times p$, with each entry $C_{ij}=\sum_k a_{ik}b_{kj}$ formed by "dotting" row $i$ of $A$ with column $j$ of $B$. Matrix multiplication behaves very differently from ordinary number multiplication: it is generally not commutative ($AB\ne BA$, and one product may not even be defined while the other is); it is associative ($(AB)C=A(BC)$) and distributes over addition ($A(B+C)=AB+AC$); multiplying by the identity matrix $I$ leaves a matrix unchanged ($AI=IA=A$); and — sharply unlike numbers — the product of two non-zero matrices can equal the zero matrix. Because $AB\ne BA$, algebraic identities that are automatic …