How can I divide by a matrix? I can multiply a vector by a matrix like so:
$$\begin{pmatrix}a\\ b \\ c\end{pmatrix}\cdot\begin{pmatrix}d & e & f \\ g & h & i \\ j & k & l \end{pmatrix} = \begin{pmatrix}  ad + be + cf \\ ag + bh + ci \\  aj + bk + al\end{pmatrix},$$
but how do I divide? 
$$\begin{pmatrix} a \\ b\\c\end{pmatrix}/\begin{pmatrix}  d &e& f\\ g& h& i \\ j& k& l\end{pmatrix}=\,?$$
I've looked everywhere and simply can't find an answer.
 A: For a matrix $$A=\begin{bmatrix}d&e&f\\g&h&i\\j&k&l\end{bmatrix}$$
and a vector $y=\begin{bmatrix}a\\b\\c\end{bmatrix}$
the product $y\cdot A$ is not defined.
The product $A\cdot y$, on the other hand, is defined by
$$Ay = \begin{bmatrix}ad + be + cf\\ag + bh + ci\\aj + bk + al\end{bmatrix}$$

But your question is how to divide by the matrix, and the answer is: it's complicated. In general, division by a matrix is not well defined and is usually not referred to as dividing by a matrix.

First, let's see division in real numbers. What does $x=\frac{a}b$ really mean? $x=\frac ab$ really means that $x$ is the unique solution to the equation
$$bx=a$$
For example, $\frac12$ is the number that we have to multiply by $2$ to get the result of $1$.

Similarly, when you want to ask "What is $y$ divided by $A$", what you are really asking is 

Which vector $x$ do I need to multiply by $A$ to get $y$?

Or, in other words, you are solving the equation $$Ax=y$$
and you want to solve it for $x$.
Now, you run into problem. 


*

*What if all elements of $A$ are $0$, but $y$ is not all zero? Then obviously, any vector $x$, multiplied by $A$, will be equal to $0$, so there is no solution to $Ax=y$.

*What if all elements of $A$ and $y$ are both zero? Then any vector $x$ will be a solution to $Ax=y$.


These two problems are similar to the division by $0$ problem in real numbers: The equation $0\cdot x=1$ has no solutions, and $0\cdot x=0$ has infinitely many solutions.
However, it gets worse. The matrix
$$A=\begin{bmatrix}1&0&0\\1&0&0\\1&0&0\end{bmatrix}$$
also causes problems, since $$A\cdot\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}a\\a\\a\end{bmatrix}$$
no matter what $b,c$ are. Again, you have zero solutions for the equation $Ax=\begin{bmatrix}1\\0\\0\end{bmatrix}$ and infinitely many solutions for $Ax=\begin{bmatrix}1\\1\\1\end{bmatrix}$.
A: If $A$ and $B$ are the matrices, and $B$ is square and invertible, then $A / B$ can be interpreted as $A$ * inverse($B$), assuming the number of columns of $A$ is the same as the number of rows of inverse($B$).
Related to this, if $B$ is square and invertible, then inverse($B$) = $B^{-1}$. So $B^{-n} = (\text{inverse}(B))^n$.
