Multiplying a matrix equation by a matrix $$A=\left[\begin{matrix}
1&2\\
-2&3\\
1&-1
\end{matrix}\right]{,}\ B=\left[\begin{matrix}
1&-2&-4\\
-1&4&9
\end{matrix}\right]{,}\ b=\left[\begin{matrix}
1\\
3\\
5
\end{matrix}\right]$$
$$ BA=I_2$$
Solving the equation  $$A x=b.$$
I start by multiplying with the matrix B.
$$ BAx=Bb \\
\left[\begin{matrix}
1&0\\
0&1
\end{matrix}\right]x=\left[\begin{matrix}
1&-2&-4\\
-1&4&9
\end{matrix}\right]\left[\begin{matrix}
1\\
3\\
5
\end{matrix}\right] =\left[\begin{matrix}
-25\\
56
\end{matrix}\right] $$
If $$x=\left[\begin{matrix}
x_1\\
x_2
\end{matrix}\right], $$ then
$$\left[\begin{matrix}
1&0\\
0&1
\end{matrix}\right]\left[\begin{matrix}
x_1\\
x_2
\end{matrix}\right]=\left[\begin{matrix}
-25\\
56
\end{matrix}\right] \\
\left[\begin{matrix}
x_1\\
x_2
\end{matrix}\right]=\left[\begin{matrix}
-25\\
56
\end{matrix}\right]. $$
Solution should be $$ x=\left[\begin{matrix}
-25\\
56
\end{matrix}\right], $$ but it's not, since
$$ Ax=\left[\begin{matrix}
1&2\\
-2&3\\
1&-1
\end{matrix}\right]\left[\begin{matrix}
-25\\
56
\end{matrix}\right]=\left[\begin{matrix}
87\\
218\\
-81
\end{matrix}\right]\ne b$$ and this equation doesn't have solutions at all.
So at what point does this go wrong? Is there some rule I don't know?
 A: I'd think about this in geometrical terms. A multiplication by a non-singular square matrix is just a change in coordinate system, and doesn't change the intrinsic nature of the equation. Therefore any solution to the equation after the multiplication is a solution to the original equation. Such a multiplication transforms the equation into an equivalent equation.
Compared to that a multiplication by a non-square matrix with fewer rows than columns is a projection onto a linear space with fewer dimensions. You lose information from the original equation. If the original equation does have a solution, then the projected version will have a projected instance of that solution, corresponding to the same variable vector $x$. But you might well get a solution as an artifact of the projection, which doesn't have a corresponding solution in the original.
To illustrate that analogy, think about the problem of intersecting two lines in 3D space. When they do intersect and you project them onto a plane, the projected lines will intersect in the projection of the original point of intersection. But you might have lines which are skew in 3d and still find an intersection in 2d.
A solution to the lower dimensional system is required for any solution of the higher dimensional system, but it is not sufficient.
You can make similar arguments for square but singular matrices, which you can think of as a projection onto a lower dimensional space followed by an embedding of that into a space of the original dimension.
