$2\times 2$ matrix equation Let $A$, $B$ and $C$ be $2 \times 2$ matrices. Let $X$ be an unknown $2 \times 2$ matrix. Consider the equation $$AX+XB = C$$
The question is: Why can you not solve this equation as $2 \times 2$ systems simultaneously? Write the matrix as four equations with four unknowns.
I tried something like:
Let $A = \begin{bmatrix}
    a_1       & a_2\\
    a_3       & a_4\\ \end{bmatrix}$ , $B = \begin{bmatrix}
    b_1       & b_2\\
    b_3       & b_4\\ \end{bmatrix}$, $C = \begin{bmatrix}
    c_1       & c_2\\
    c_3       & c_4\\ \end{bmatrix}$
 and $X = \begin{bmatrix}
    x_1       & x_2\\
    x_3       & x_4\\ \end{bmatrix}$
$AX = \begin{bmatrix}
    a_1x_1+a_2x_3       & a_1x_2+a_2x_4\\
    a_3x_1+a_4x_3       & a_3x_2+a_4x_4\\ \end{bmatrix}$
$XB = \begin{bmatrix}
    x_1b_1+x_2b_3       & x_1b_2+x_2b_4\\
    x_3b_1+x_4b_3       & x_3b_2+x_4b_4\\ \end{bmatrix}$
And thus
$$AX + XB = \begin{bmatrix}
    a_1x_1+a_2x_3+x_1b_1+x_2b_3       & a_1x_2+a_2x_4+x_1b_2+x_2b_4\\
    a_3x_1+a_4x_3+x_3b_1+x_4b_3       & a_3x_2+a_4x_4+x_3b_2+x_4b_4\\ \end{bmatrix} = \begin{bmatrix}
    c_1       & c_2\\
    c_3       & c_4\\ \end{bmatrix}$$
I'm not really sure what I ended up with here, I guess if we assume that $A, B, C$ are known  we do have 4 equations with 4 unknowns (just set the first entry in AX+XB equal to $c_1$, second entry $c_2$ and so forth... Is this the correct solution? Is there an easy way to explain why we can't solve it as a double system simultaniously?
 A: If you know $A,B,$ and $C$, then a solution to $AX + XB = C$ is a solution to the following equation, and vice versa:
$$
\begin{bmatrix}
(a_1 + b_1) & b_3 & a_2 & 0 \\
b_2 & (a_1 + b_4) & 0 & a_2 \\
a_3 & 0 & (a_4 + b_1) & b_3 \\
0 & a_3 & b_2 & (a_4 + b_4)
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{bmatrix} = 
\begin{bmatrix}
c_1 \\
c_2 \\
c_3 \\
c_4 \\
\end{bmatrix}.$$
This reduces to matrix computation we know well. 
A: You can solve the matrix equation if $ A $ and $ B $ satisfy some appropriate hypotheses. For example,


*

*Supose that $BX=XB$ and exist $S$ such that $S(A+B)=I$. Then 
\begin{align}
AX + XB = & C
\\
AX + B X= & C
\\
(A+B)X = & C
\\
S(A+B)X = &S C
\\
X = &SC
\\
\end{align}

*Supose that $AX=XA$ and exist $R$ such that $(A+B)R=I$. Then 
\begin{align}
AX + XB = & C
\\
XA +  XB= & C
\\
X(A+B) = & C
\\
X(A+B)R = & CR
\\
X = &CR
\\
\end{align}

*Suppose $AX=XU$ for some $U$ and exist $R$ such that $(U+B)R=I$. Then 
\begin{align}
AX + XB = & C
\\
XU + B X= & C
\\
X(U+B) = & C
\\
X(U+B)R = & CR
\\
X = &CR
\\
\end{align}

*Supose that $VX=XB$ and for some $V$ and exist $S$ such that $S(A+V)=I$. Then 
\begin{align}
AX + XB = & C
\\
XA +  VX= & C
\\
(A+V)X = & C
\\
S(A+B)X = & SC
\\
X = &SC
\\
\end{align}
