How to find matrix multiplications like AB = 10A+B? saw this one.

How do we find others (not necessarily 2 by 2)?
How do we generalize it?
 A: For any matrix $A$ you can find a matrix $B$ such that
$$ A\,B = \lambda A +B $$
Just solve the equation algebraically for $B$
$$\begin{aligned}A\,B & =\lambda A+B\\
A\,B-B & =\lambda A\\
\left(A-{\bf 1}\right)B & =\lambda A\\
B & =\lambda\left(A-{\bf 1}\right)^{-1}A
\end{aligned}$$
where ${\bf 1}$ is the appropriately sized identity matrix, and $\lambda = 10$ is the specific factor in this problem.
The requirement seems to be that $A-1$ is invertible.
A: To multiply matrix A by matrix B, think of the ith row of matrix A as a vector, $u_i$, and the jth column of matrix B as a vector, $v_j$.  Then $(AB)_{ij}$, the number in the ith row, jth column is the "dot product" of those two vectors.
For example, to multiply $AB= \begin{pmatrix}3 & 4 \\ 8 & 7 \end{pmatrix}\begin{pmatrix} 7 & 2 \\ 4 & 9 \end{pmatrix}$, the first row of A is the vector <3 4> and the first column of B is the vector <7 4>.  Their dot product is 3(7)+ 4(4)= 21+ 16= 37. $(AB)_{11}= 37$
The first row of A is the vector <3, 4> and the second column of B is the vector <2, 9>.  Their dot product is 3(2)+ 4(9)= 6+ 36= 42.  $(AB)_{12}= 42$.
The second row of A is the vector <8, 7> and the first column of B is the vector <7, 4>.  Their dot product is 8(7)+ 7(4)= 56+ 28= 84.  $(AB)_{21}= 84$.
The second row of A is the vector <8, 7> and the second column of B is the vector <2, 9>.  Their dot product is 8(2)+ 7(9)= 16+ 63= 79.  $(AB)_{22}= 79$.
You can do the same thing with higher dimension matrices.
