Suppose we have the following matrix $$ C = \begin{bmatrix} a_{11} u B^{\top}_{\bullet1}x & a_{21} u B^{\top}_{\bullet1}x\\ a_{21} v B^{\top}_{\bullet2}x & a_{22} v B^{\top}_{\bullet2}x \end{bmatrix} $$ where $B$ is a matrix with two columns and number of rows equal to the element of $x$. Also, $B^{\top}_{\bullet1}$ is the transpose of the first column of $B$.
Would it be possible to decompose $C$ into $4$ different matrices as $A, B$, matrix including $u,v$(which are scalars), and $x$ by not copying columns or rows of either of these matrices.
My try: I can write the following
$$ C = \begin{bmatrix} a_{11} u B^{\top}_{\bullet1}x & a_{21} u B^{\top}_{\bullet1}x\\ a_{21} v B^{\top}_{\bullet2}x & a_{22} v B^{\top}_{\bullet2}x \end{bmatrix} = \begin{bmatrix} u & 0\\ 0 & v \end{bmatrix} \begin{bmatrix} B^{\top}_{\bullet1}x & 0\\ 0 & B^{\top}_{\bullet2}x \end{bmatrix} \begin{bmatrix} a_{11} & a_{21} \\ a_{21} & a_{22} \end{bmatrix} $$
I want to have $x$ using a matrix product not within another matrix. Maybe my way is not the good start.
I assume that $B^{\top}_{\bullet1}$ is the transpose of $B^{}_{\bullet1}$, which is the first column of $B$. If that is the case, then your matrix can be written as $$ \pmatrix{a_{11} & a_{21}\\a_{21} & a_{22}}\pmatrix{u&0\\ 0&v} B^Tx. $$