Using the Left-Inverse to "Solve" an Impossible System of Equations I was working with the following system of equations:
$$\begin{split}
    \begin{bmatrix}
    4 & 0\\
    0 & 5\\
    0 & 0\\
    \end{bmatrix}
    \begin{bmatrix}
    x_1\\
    x_2\\
    \end{bmatrix}
     & = 
    \begin{bmatrix}
    1\\
    1\\
    1\\
    \end{bmatrix}
    \end{split}
$$
Clearly, this has no solution on account of the last rows in the coefficient and solution matrices. 
However, multiplying by the left-inverse of the coefficient matrix seems to imply a solution:
$$\begin{split}
    \begin{bmatrix}
    \frac{1}{4} & 0 & b_{13}\\
    0 & \frac{1}{5} & b_{23}\\
    \end{bmatrix}
    \begin{bmatrix}
    4 & 0\\
    0 & 5\\
    0 & 0\\
    \end{bmatrix}
    \begin{bmatrix}
    x_1\\
    x_2\\
    \end{bmatrix}
     & = 
    \begin{bmatrix}
    \frac{1}{4} & 0 & b_{13}\\
    0 & \frac{1}{5} & b_{23}\\
    \end{bmatrix}
    \begin{bmatrix}
    1\\
    1\\
    1\\
    \end{bmatrix} \\
    \begin{bmatrix}
    x_1\\
    x_2\\
    \end{bmatrix}
     & = 
    \begin{bmatrix}
    \frac{1}{4} + b_{13}\\
    \frac{1}{5} + b_{23}\\
    \end{bmatrix}
    \end{split}
$$
Where $b_{13}$ and $b_{23}$ can be any number. As stated above, there is no solution to this system, so any solution obtained by the above method is wrong.
What I am not understanding is why, after applying the rules of matrix multiplication, does it seem possible that there is a solution? Did I make a mistake somewhere in the process, and/or is there some fundamental nuance of linear algebra that I'm missing here?

Here is a simpler example of what you are doing. Consider the argument
\begin{align}
   2 \times 3 &= 8 \\
   \frac 12 \times 2 \times 3 &= \frac 12 \times 8 \\
   3 &= 4
\end{align}
If you put $``="$ between two quantities that aren't equal, then you cannot believe any conclusions that may follow.
\begin{equation}
    \begin{split}
       \begin{bmatrix}
          4 & 0\\
          0 & 5\\
          0 & 0\\
       \end{bmatrix}
       \begin{bmatrix}
          x_1\\
          x_2\\
       \end{bmatrix}
       & = 
       \begin{bmatrix}
          1\\
          1\\
          1\\
       \end{bmatrix}
    \end{split}
\end{equation}
Is equivalent to
\begin{equation}
    \begin{split}
       \begin{bmatrix}
          4x_1\\
          5x_2\\
          0\\
       \end{bmatrix}
       & = 
       \begin{bmatrix}
          1\\
          1\\
          1\\
       \end{bmatrix}
    \end{split}
\end{equation}
Which clearly has no solution.
Geometrically, the point $(1,1,1)$ is not on the plane spanned by the vectors $(4,0,0)$ and $(0,5,0)$.
Equations like this occur all of the time in statistics. Their, $``$solution$"$ would be to add an error term on the right end
$$\left[
   \begin{array}{c}
      4 & 0\\
      0 & 5\\
      0 & 0\\
   \end{array}
\right]
\left[
   \begin{array}{c}
      x_1\\
      x_2\\
   \end{array}
\right]
= 
\left[
   \begin{array}{c}
      1\\
      1\\
      1\\
   \end{array}
\right]
+
\left[
   \begin{array}{c}
      \epsilon_1\\
      \epsilon_2\\
      \epsilon_3\\
   \end{array}
\right]$$
and then find the values of $x_1$ and $x_2$ that will minimize the sum of the squares of the error: $\epsilon_1^2 + \epsilon_2^2 + \epsilon_3^2$ 
Geometrically, that would be the perpendicular projection of the point
$(1,1,1)$ onto the plane spanned by the vectors $(4,0,0)$ and $(0,5,0)$.
It turns out that, even if $Ax = b$ has no solution, $A^TAx = A^Tb$ does have a solution, namely $\hat x = \left(A^TA \right)^{-1}A^Tb$, and $\hat x$ minimizes the sum of the squares of the errors, $\|b- A\hat x \|^2$
