Why can't you swap rows in the matrix for a system of linear differential equations? If you are given a Matrix A, and then asked to solve the initial value problem x'=Ax, why can one not swap rows before starting the problem.  I tried it with a 3x3 matrix on wolfram alpha and got two different results, but what is the reason behind this?
 A: Below I've used the numbers $1,\ldots,9$ in $A$ because it's easier to understand (and to write), but there's no loss of generality.
Here is the original case
$$\begin{align}
\left[\array{x_1'\\x_2'\\x_3'}\right]
&= \left[\array{1 & 2 & 3\\4&5&6\\7&8&9}\right]
\left[\array{x_1\\x_2\\x_3}\right]\\
&\iff\\
x_1'&=1x_1+2x_2 + 3x_3\\
x_2'&=4x_1+5x_2 + 6x_3\\
x_3'&=7x_1+8x_2 + 9x_3\\
\end{align}$$
Now swap the first and second row and what you find is that you have different equations for $x_1'$ and $x_2'$
$$\begin{align}
\left[\array{x_1'\\x_2'\\x_3'}\right]
&= \left[\array{4&5&6\\1 & 2 & 3\\7&8&9}\right]
\left[\array{x_1\\x_2\\x_3}\right]\\
&\iff\\
x_1'&=4x_1+5x_2 + 6x_3\\
x_2'&=1x_1+2x_2 + 3x_3\\
x_3'&=7x_1+8x_2 + 9x_3\\
\end{align}$$
But if you also swap the rows in the vector $\mathbf{x'}$ you are ok
$$\begin{align}
\left[\array{x_2'\\x_1'\\x_3'}\right]
&= \left[\array{4&5&6\\1 & 2 & 3\\7&8&9}\right]
\left[\array{x_1\\x_2\\x_3}\right]\\
&\iff\\
x_2'&=4x_1+5x_2 + 6x_3\\
x_1'&=1x_1+2x_2 + 3x_3\\
x_3'&=7x_1+8x_2 + 9x_3\\
\end{align}$$
In terms of linear algebra, what we have done in the last, correct case is pre-multiplied both sides of your original equation $\mathbf{x'}=A\mathbf{x}$ by $P$
$$\mathbf{x'}=A\mathbf{x} \;\;\;\text{equivalent to}\;\;\; P\mathbf{x'}=PA\mathbf{x} $$
where
$$ P= \left[\array{
0&1&0\\
1 & 0 & 0\\
0&0&1}\right]
$$
In the incorrect case we only pre-multiplied the RHS of the equation by P.
$$\mathbf{x'}=A\mathbf{x} \;\;\text{not equivalent to}\;\;\; \mathbf{x'}=PA\mathbf{x} $$
A: When you swap rows, you are re-arranging the variables. So you also have to swap columns since each column goes with a variable.
