Proof two matrices equation can be combined

Question

If $A \begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$ and $A \begin{pmatrix}4\\6\end{pmatrix}=\begin{pmatrix}0\\2\end{pmatrix}$ then $A$ is...

The author said that I can "combine" those equations to $$A \begin{pmatrix}2 & 4\\1 & 6\end{pmatrix}=\begin{pmatrix}1&0\\0&2\end{pmatrix}$$

My question is how to prove that I can combine those equations. Here's my attempt to prove

Let $A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$ and I get four linear equations $$2a+b=1\\2c+d=0\\4a+6d=0\\4c+6d=2$$ Make it to an augmented matrix $$\left[\begin{array}{rrrr|r} 2 & 1 & 0 & 0 & 1 \\ 4 & 6 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0\\ 0 & 0 & 4 & 6 & 2 \end{array}\right]$$ I have no clue what to do next

Answer 1

$A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$

And you conclude: $$ 2a+b=1\\2c+d=0\\4a+6d=0\\4c+6d=2 $$

We get:

$$ \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}2 & 4\\1 & 6\end{pmatrix}=\begin{pmatrix}2a+b & 4a+6b\\2c+d&4c+6d\end{pmatrix} = \begin{pmatrix}1&0\\0&2\end{pmatrix} $$

Answer 2

This can be understood by going back to the basic definition of the product of an $m\times r$ matrix $A$ with an $r\times n$ matrix $B$: $$(AB)_{ij} = \sum_{k=1}^r a_{ik}b_{kj}.$$ Let’s examine the first column of this product. Applying the above definition element by element it’s going to be $$\pmatrix{\sum_{k=1}^r a_{1k}b_{k1} \\ \sum_{k=1}^r a_{2k}b_{k1} \\ \vdots \\ \sum_{k=1}^r a_{mk}b_{k1}} = A \pmatrix{b_{11} \\ b_{21} \\ \vdots \\ b_{r1}}$$ That is, it’s just the product of $A$ with the first column of $B$. Generalizing, we can see that the $j$th column of $AB$ is equal to $A$ times the $j$th column of $B$.

Using this idea, we can combine the two equations into a single matrix equation by “gluing together” the vectors in the individual equations.