In p.279-280 of "Conceptual Mathematics" 2nd ed. the book talks about maps from coproducts to products. For any map $f: A + B \rightarrow X \times Y$, $f$ can be represented as a matrix of maps from the summands to the factors:
$$ \begin{bmatrix}A \rightarrow X&A \rightarrow Y\\B \rightarrow X&B \rightarrow Y\end{bmatrix} $$
The book then discusses the idea of linear categories which have for any two objects $X$ and $Y$, a special map $0_{XY}$ such that this map composed with any other maps gives another $0$ map. Specifically, linear categories have the property that the following "identity matrix" $X + Y \rightarrow X \times Y$ is an isomorphism:
$$ \begin{bmatrix}1_{X}&0_{XY}\\0_{YX}&1_{Y}\end{bmatrix} $$
With this we can then "multiply matrices" $f: A + B \rightarrow X \times Y$ and $g: X + Y \rightarrow U \times V$ via..
$$ \begin{bmatrix}f_{AX}&f_{AY}\\f_{BX}&f_{BY}\end{bmatrix} \cdot \begin{bmatrix}g_{XU}&g_{XV}\\g_{YU}&g_{YV}\end{bmatrix} = \begin{bmatrix}g_{XU}&g_{XV}\\g_{YU}&g_{YV}\end{bmatrix} \circ \begin{bmatrix}1_{X}&0_{XY}\\0_{YX}&1_{Y}\end{bmatrix}^{-1} \circ \begin{bmatrix}f_{AX}&f_{AY}\\f_{BX}&f_{BY}\end{bmatrix}$$
which is
$$ A + B \xrightarrow[]{f} X \times Y \xrightarrow[]{\alpha} X + Y \xrightarrow[]{g} U \times V$$
where $\alpha$ is the inverse of the corresponding identity matrix.
With that in place we can also "add" two maps $f,g: A \rightarrow B$ determined by:
$$ \begin{bmatrix}1_{AA}&f\\0_{BA}&1_{BB}\end{bmatrix} \cdot \begin{bmatrix}1_{AA}&g\\0_{BA}&1_{BB}\end{bmatrix} = \begin{bmatrix}1_{AA}&f + g\\0_{BA}&1_{BB}\end{bmatrix}$$
The question is then:
Using the above definitions of matrix multiplication and addition of maps, prove the following formula for matrix multiplication:
$$ \begin{bmatrix}f_{AX}&f_{AY}\\f_{BX}&f_{BY}\end{bmatrix} \cdot \begin{bmatrix}g_{XU}&g_{XV}\\g_{YU}&g_{YV}\end{bmatrix} = \begin{bmatrix}g_{XU} \circ f_{AX} + g_{YU} \circ f_{AY}&g_{XV} \circ f_{AX} + g_{YV} \circ f_{AY}\\g_{XU} \circ f_{BX} + g_{YU} \circ f_{BY}&g_{XV} \circ f_{VX} + g_{YV} \circ f_{BY}\end{bmatrix} $$
I've attempted this by starting with the upper left sum $g_{XU} \circ f_{AX} + g_{YU} \circ f_{AY}$ and using the equation for sums of maps, which got me to
$$\begin{bmatrix}1_{AA}&g_{YU} \circ f_{AY}\\0_{UA}&1_{UU}\end{bmatrix} \circ \begin{bmatrix}1_{A}&0_{AU}\\0_{UA}&1_{U}\end{bmatrix}^{-1} \circ \begin{bmatrix}1_{AA}&g_{XU} \circ f_{AX}\\0_{AU}&1_{UU}\end{bmatrix}$$
but I am not sure where to go from here. My first instinct is the above matrix is composed position-wise to get something like
$$\begin{bmatrix}1_{A}&g_{YU} \circ f_{AY} \circ 0_{AU}^{-1} \circ g_{XU} \circ f_{AX}\\0_{UA}&1_{U}\end{bmatrix}$$
which then matches position-wise with the definition of matrix multiplication would imply..
$$ g_{YU} \circ f_{AY} \circ 0_{AU}^{-1} \circ g_{XU} \circ f_{AX} = g_{XU} \circ f_{AX} $$
which in turn implies..
$$ g_{YU} \circ f_{AY} \circ 0_{AU}^{-1} = 1_{U} $$
and therefore
$$ g_{YU} \circ f_{AY} = 0_{AU} $$ ?
But then how do I prove that?
Any help would be appreciated, thanks!