How do I solve this matrix equation? How do I solve this matrix equation and what is the answer:
$$\begin{bmatrix}
  -122.366667 \\
  37.61666667
\end{bmatrix} = \begin{bmatrix}
  0.000046 & 0.000032 & -122.413307 \\
  0.000025 & -0.000036 & 37.632195
\end{bmatrix}\begin{bmatrix}
  x \\
  y \\
  1
\end{bmatrix}$$
 A: In general, to solve for $C$ when you have $A = BC$ and they're all matrices, you have to:


*

*find the inverse of $B$, call it $B^{-1}$

*left-multiply each side by $B^{-1}$ to get $B^{-1}A = C$

A: We can write out the matrix multiplication to get a system of equations, and then solve that. Here is the expanded matrix multiplication: 
$$
0.000046x + 0.000032y - 122.413307 = -122.366667
$$$$
0.000025x - 0.000036y + 37.632195 = 37.61666667
$$
So we end up with two equations of two variables, which can be easily solved by a method of your choice (substitution, elimination, matricies, etc).
$$
x\approx 481.325,   y\approx 765.596
$$
as you should verify.
A: Note that, this can be brought down to the following square system since 1 is not a variable hence can be included at the left hand side. (Just multiply everything to see this.) 
$$\pmatrix{
  122.413307-122.366667 \\
  -37.632195 + 37.61666667
} = \pmatrix{0.046640\\ -0.01552833}=\begin{bmatrix}
  0.000046 & 0.000032\\
  0.000025 & -0.000036 \end{bmatrix}\pmatrix{
  x \\
  y}
$$
Now you are back at the $b=Az$ form which can be solved by multiplying both sides by the matrix $A^{-1}$ to obtain $z=A^{-1}b$. Again, this only makes sense if your matrix $A$ is invertible if not you have to apply further steps. 
$$
z = A^{-1}b = \pmatrix{
          14657.9804560261         & 13029.3159609121\\
          10179.1530944625         &-18729.6416938111
}\pmatrix{0.046640\\ -0.01552833} \approx \pmatrix{481.3
\\765.6}
$$
I did not pay attention to the significant digits (and it felt good!) hence you have to take care of that.
