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I want to solve an equation with Lagrange multiplier. I know elimination method but I need to know this method. I wonder if anyone let me know how to estimate $\mu$ and 3 unknown variables ($\hat A$) algebraically .

$n.\hat\mu$ + $n_1\hat A_1$ + $n_2\hat A_2$ + $n_3 \hat A_3$ + $0 \lambda$ = $y_{..}$
$n_1\hat\mu$ + $n_1\hat A_1$ + $1 \lambda$ = $y_{1.}$
$n_2\hat\mu$ + $n_2\hat A_2$ + $1 \lambda$ = $y_{2.}$
$n_3\hat\mu$ + $n_3 \hat A_3$ + $1 \lambda$ = $y_{3.}$
$0\hat\mu$ + $1\hat A_1$ + $1\hat A_2$ + $1 \hat A_3$ + $0 \lambda$ = $0$

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This is 5 linear equations in 5 unknowns.

So ... you say you know "the elimination method" by which I assume you mean Gaussian elimination, which should work.

What's the problem. You know a way to solve such a set of equations, and you have the equations you need. If you know the statistics (the $n$'s and $y$'s in your formulae), you can immediately apply the approach you already know.

Is there a numerical issue?

Or is the problem that you want to solve these equations algebraically??

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  • $\begingroup$ I want to solve these eq. algebraically. In Gaussian elimination, I use constraint to solve the 4 main LS equations, for example equals $\mu=0$ and estimate 3 unknowns. In this case all solutions will not be estimable. To solve the problem of estimablity, I discovered it is possible to use Lagrange multiplier. But now I don't know how it works. $\endgroup$ – mohammad May 10 '14 at 10:10
  • $\begingroup$ Are $n1$ and $n_1$ intended to be the same thing? If so, can you edit your question, please, to fix that? Can you also edit to reflect your actual needs - it's important to mention things like a wish for an algebraic answer. $\endgroup$ – Glen_b -Reinstate Monica May 10 '14 at 10:14

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