Find matrices A and B I am coming across questions like these:
Find $A$ and $B$ if


*

*$$2A+3B=I_2$$
$$ A+B=2A^t$$

*$$2A+3B=  \left( \begin{array}{cc}
8 & 3 \\
7 & 6 \end{array} \right)$$
$$A+B^t= \left( \begin{array}{cc}
3 & 1 \\
3 & 3 \end{array} \right)$$


*$$2A+B^t= \left( \begin{array}{cc}
2 & 5 \\
10 & 2 \end{array} \right)$$
$$2B+A^t= \left( \begin{array}{cc}
1 & 8 \\
4 & 1 \end{array} \right)$$


Problems like these haven't been done in my books and I couldn't find anything online either. Can someone please give me some hints or any links about how to do these types of problems?
 A: You can just use 
$$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$$
$$B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$$
and then solve the system of equations for $a_{ij}, b_{ij}$. For each matrix-equation you will get 4 scalar equations (one for each entry).
You can also try to simplify the equations in a more direct way with algebraic manipulations, 
E.g: Third example: Transpose the second equation, then multiply it by -2 and add it to the first one. Then you will get (IF I've got the calculations right=):
$$ -5B^t = \begin{pmatrix} -2 & -11 \\ -22 & -2\end{pmatrix}$$ which you can now solve easily for $B$ and plug it into one of the equations for solving it for $A$.
A: For 1, multiply the second equation by $3$ and subtract the first:
$$
A=6A^t-3I_2
$$
Now transpose: $A^t=6A-3I_2$, so
$$
A=6(6A-3I_2)-3I_2
$$
or
$$
35A=21I_2
$$
Once you know $A$, it's easy to compute $B$ from the first equation.
For 2, transpose the second equation and eliminate $B$, then do similarly.
A: take elements of $A$ as $x_1,x_2,x_3,x_4$ and $B$ as $y_1,y_2,y_3,y_4$
substitute them in the equaions and also find  $A^T$
after simplifying 
then compare the rows and columns.
you will get equations solve it and get the results of $x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4$  and there you find $A$ and $B$
