How to find matrix multiplications like AB = 10A+B?

Question

saw this one.

How do we find others (not necessarily 2 by 2)? How do we generalize it?

Answer 1

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.

Answer 2

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.