Solving matrix equation $AX = B$ I want to solve the matrix equation $AX = B$, where the matrix $A$ and $B$ are given as follows 
$A = 
  \begin{bmatrix}
     0.1375 &   0.0737  &  0.1380  & 0.1169 &  0.1166 \\
    0.0926  &   0.0707   &  0.0957   &  0.0873  &   0.0733 \\
    0.0767   &   0.0642  &   0.0810    &  0.0766  &    0.0599 \\
    0.1593 &    0.1020 &    0.1636 &    0.1451   &  0.1317
  \end{bmatrix}$
$B = 
  \begin{bmatrix}
     0.2794   &   0.0065  &    0.2271    &  0.1265   &   0.2773\\
    0.1676  &  0.2365  &  0.1430  &  0.1015 &   0.0632 \\
     0.0645  &   0.2274 &    0.1009 &    0.1806 &    0.0503\\
    0.2326  &  0.1261  &  0.2867 &   0.2846  &  0.1979
  \end{bmatrix}$
Could anybody help me how to solve this problem? I need help with this.
Thanks for the help.
 A: Write $B=(b_1,\dots,b_n)$ and $X=(x_1,\dots,x_n)$ with $b_i$ and $x_i$ the columns of $B$ and $X$ respectively. I assume you are able to solve $Ax=b$. Now, solve the $n$ linear equations $Ax_1=b_1$ to $Ax_n=b_n$. You can now 'glue' the $x_i$ together to get the matrix $X$.
A: A little theory: It is concerned with the Kronecker product and the vectorization of matrices.
The Kronecker product, denoted by $\otimes$, is an operation on two matrices of arbitrary size resulting in a block matrix. For example, $A \in R^{m \times n}, B \in R^{p \times q}$, then the Kronecker product $A \otimes B$ is the $mp \times nq$ block matrix:
$\begin{bmatrix} a_{11}B & \cdots & a_{1n}B \\ \cdots & \cdots & \cdots \\ a_{m1}B & \cdots & a_{mn}B  \end{bmatrix}$.
The vectorization of a matrix $X$, denoted by $\operatorname{vec}(X)$, is formed by stacking the columns of $X$ into a single column vector. For example, if $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$, then $\operatorname{vec}(A) = \begin{bmatrix} 1 \\ 4 \\ 2 \\ 5 \\ 3 \\ 6 \end{bmatrix}$.
The key point here is:

The Kronecker product, together with the vectorization, can be used to get a convenient representation for some matrix equations. For the matrix equation $AXB = C$, where $A, B$ and $C$ are given matrices and the matrix $X$ is the unknown. We can rewrite this equation as:
   $(B^\top \otimes A) \operatorname{vec}(X) = \operatorname{vec}(AXB) = \operatorname{vec}(C)$. Now, we get a linear system and you can solve it in standard ways. 

For your problem: $AX = B \Rightarrow AXI = B \Rightarrow \operatorname{vec}(AXI) = \operatorname{vec}(B) \Rightarrow (I^{\top} \otimes A) \operatorname{vec}(X) = \operatorname{vec}(B)$ $\Rightarrow (I \otimes A) \operatorname{vec}(X) = \operatorname{vec}(B)$. It is a linear system now.
There are also some useful theorems on Kronecker product to help you to identify when your matrix equation has a unique solution.
