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How to transform the equation $\frac{\partial L(m,\lambda)}{\partial m}=2 \cdot \Sigma m+ \lambda \cdot \mathbf{1}$ into $m=-\frac{1}{2} \cdot \lambda \Sigma^{-1} \mathbf{1}$, where $\mathbf{1} = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix}\ \right), m=(m_a,m_b,m_c), \Sigma$ is a $3 \times 3$ matrix, $\lambda$ is a scalar?

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closed as unclear what you're asking by BLAZE, Servaes, SchrodingersCat, Crostul, user223391 Nov 29 '15 at 5:20

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This holds if and only if $\frac{\partial L}{\partial m} = 0$. In this case, you obtain the equation \begin{equation} 0 = 2\cdot\Sigma m + \lambda\cdot \mathbf{1}, \end{equation} which is equivalent with your expression for $m$ after shuffeling some terms around.

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