Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
1  
This doesn't seem to be a matrix-vector equation of the conventional sort... –  J. M. 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. Oct 26 '11 at 23:30
    
try j.gs/2Btq you get quick answers –  chndn May 20 '13 at 15:17
add comment

3 Answers

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.

share|improve this answer
    
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
add comment

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 $B^{-1}A = C$
share|improve this answer
    
And if $B$ isn't invertible? –  user5137 Oct 27 '11 at 0:38
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.