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How do I solve this matrix equation and what is the answer:

$$\begin{bmatrix} -122.366667 \\ 37.61666667 \end{bmatrix} = \begin{bmatrix} 0.000046 & 0.000032 & -122.413307 \\ 0.000025 & -0.000036 & 37.632195 \end{bmatrix}\begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$$

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This doesn't seem to be a matrix-vector equation of the conventional sort... –  J. M. is back. Oct 26 '11 at 23:18
Sorry, I am new to this. I have adjusted the syntax. –  Andrew Johnson Oct 26 '11 at 23:25
You can try striking out the last column of the coefficient matrix and last row of the variable vector, yielding a usual 2-by-2 system. See if the solution of that system satisfies your original system... –  J. M. is back. Oct 26 '11 at 23:30
try you get quick answers –  chndn May 20 '13 at 15:17

3 Answers 3

up vote 1 down vote accepted

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.

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You seems to have lost the decimal point in -122.366667. This may make a difference. –  Henry Oct 26 '11 at 23:50
@Henry many thanks, corrected. This is why it's important to verify answers :) –  smackcrane Oct 27 '11 at 0:00

In general, to solve for $C$ when you have $A = BC$ and they're all matrices, you have to:

  1. find the inverse of $B$, call it $B^{-1}$
  2. left-multiply each side by $B^{-1}$ to get $B^{-1}A = C$
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And if $B$ isn't invertible? –  user5137 Oct 27 '11 at 0:38

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.

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