Hint:
test that: $$ A(B\vec x)=(AB) \vec x $$ Using for $\vec x$ the canonical basis. It is easy for transformations in a $2$ dimensional space and requaire a bit more work for an $n-$ dimensionale space.
Given $$ A=\begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{bmatrix} \qquad B=\begin{bmatrix} b_{11}&b_{12}\\ b_{21}&b_{22} \end{bmatrix} $$ By row-column multiplication we have: $$ A(B\vec x)= \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{bmatrix} \left( \begin{bmatrix} b_{11}&b_{12}\\ b_{21}&b_{22} \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} \right)= \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{bmatrix} \begin{bmatrix} b_{11}x+b_{12}y\\ b_{21}x+b_{22}y \end{bmatrix}= $$ $$ = \begin{bmatrix} a_{11}(b_{11}x+b_{12}y)+a_{12}(b_{21}x+b_{22}y)\\ a_{21}(b_{11}x+b_{12}y)+a_{22}(b_{21}x+b_{22}y) \end{bmatrix} $$ That, reordering becomes: $$ \begin{bmatrix} (a_{11}b_{11}+a_{12}b_{21})x+(a_{11}b_{12}+a_{12}b_{22})y\\ (a_{21}b_{11}+a_{22}b_{21})x+(a_{21}b_{12}+a_{22}b_{22})y \end{bmatrix}=(AB)\vec x $$