Can I divide a matrix by a scalar I have problem where it says to find $A$. 
$$
\begin{pmatrix}
6 & 3 \\
0 & -3
\end{pmatrix} = 3A^T
$$
Can I just divide the matrix by $3$ and get 
$$
\begin{pmatrix}
2 & 1 \\
0 & -1 
\end{pmatrix} = A^T
$$
So that $A$ would be
$$
\begin{pmatrix}
2 & 0 \\
1 & -1
\end{pmatrix}
$$ ?
 A: Your solution is essentially correct, but a slight change of phrasing might make it a bit easier to understand.  Matrices have a particular kind of algebraic structure in which it makes sense to talk about scalar multiplication (in particular, if I recall the terminology correctly, $n\times n$ matrices with entries in $\mathbb{R}$ form an algebra over $\mathbb{R}$).  In particular, this means that if $A$, $B$ are any two algebra elements (i.e. any $n\times n$ matrices with real entries) and $\lambda$ is any nonzero scalar (i.e. nonzero element of the base field, which is $\mathbb{R}$ in this case), then we have
$$ A = B \iff \lambda A = \lambda B.  $$
This is the law of multiplicative cancellation, if you need a name for it.  It essentially says that we can multiply both sides of an equation by any nonzero scalar we like.
In your example, both
$$ \begin{pmatrix}
6 & 3 \\
0 & -3
\end{pmatrix}
\qquad\text{and}\qquad
3A^{T}
$$
are elements of the matrix algebra, and so
$$\begin{pmatrix}
6 & 3 \\
0 & -3
\end{pmatrix}
=
3A^{T}
\iff
\frac{1}{3}\begin{pmatrix}
6 & 3 \\
0 & -3
\end{pmatrix}
= \frac{1}{3}(3A^T).
$$
After simplifying, this reduces to
$$\begin{pmatrix} 2 & 1 \\ 0 & -1\end{pmatrix}
= A^T
\implies
A = \begin{pmatrix} 2 & 0 \\ 1 & -1 \end{pmatrix},
$$
which is exactly the result you got.  Huzzah!
