1. ## Solving matrix equations

Is there any way to solve matrix equations A=BX, where A, B and X are all n*n matrices, and you know A and B but not X, and det(A)=det(B)=0, so you can't use an inverse?

If there's no generalized way, is there any theory about how to get started on this problem? Any links or references would be appreciated.

Google can't seem to focus on this particular topic because anytime you search for solving matrix equations, you get links to solving Ax=b, where x and b are vectors and not matrices.

Thanks much!

2. Originally Posted by phileas
Is there any way to solve matrix equations A=BX, where A, B and X are all n*n matrices, and you know A and B but not X, and det(A)=det(B)=0, so you can't use an inverse?

If there's no generalized way, is there any theory about how to get started on this problem? Any links or references would be appreciated.

Google can't seem to focus on this particular topic because anytime you search for solving matrix equations, you get links to solving Ax=b, where x and b are vectors and not matrices.

Thanks much!

Try looking for decompostion of matrices.
LU decomposition.
Upper & Lower trianglular matrix manipulation.

3. From what I understand, you are looking to solve a matrix equation like this,

$\begin{bmatrix}a_{11} & a_{12} \\a_{21} & a_{22}\\ \end{bmatrix}
x=\begin{bmatrix}b_{11} & b_{12} \\b_{21} & b_{22} \\ \end{bmatrix}$

yes/no?

We can creat two equations from this...

Equation 1
$\begin{bmatrix}a_{11}\\a_{21}\\ \end{bmatrix}x_{1}+\begin{bmatrix}a_{12}\\a_{22}\\ \end{bmatrix}x_{2}=\begin{bmatrix}b_{11}\\b_{21}\\ \end{bmatrix}$

and...

Equation 2
$\begin{bmatrix}a_{11}\\a_{21}\\ \end{bmatrix}x_{1}+\begin{bmatrix}a_{12}\\a_{22}\\ \end{bmatrix}x_{2}=\begin{bmatrix}b_{12}\\b_{22}\\ \end{bmatrix}$

Solving for equation 1 we get...

$\begin{bmatrix}x_{11} & ...\\x_{21} & ...\\ \end{bmatrix}$

Solving for equation 2 we get...

$\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\\ \end{bmatrix}$

Which is the answer. Sorry, If I have over simplified your problem. This is just what I got form it.