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I have the following matrix equality: $\left( \begin{array}{ccc} u_{11} & u_{12} & -p_1 \\ u_{21} & u_{22} & -p_2 \\ p_1 & p_2 & 0 \end{array} \right).\left( \begin{array}{c} \text{dc}_1 \\ \text{dc}_2 \\ \text{d$\lambda $} \end{array} \right)=\left( \begin{array}{c} 0 \\ 0 \\ \text{dI} \end{array} \right)$

I need to solve for $\frac{\text{dc}_1 }{\text{dI}},\frac{ \text{dc}_2}{\text{dI}},\frac{\text{d$\lambda $}}{\text{dI}}$.

Attempt: I tried using Cramer's rule, but all I get is the solution for $dc_1,dc_2,d\lambda$. I need to find the solution for the $dc_1,dc_2,d\lambda$ over $dI$. Any hints please.

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    Evaluate the determinant and you will get that $dc_1 = (\text{some function depending on }u_{11},u_{12},u_{21},u_{22},p_1,p_2) dI$. Hence, you can find $\dfrac{dc_1}{dI}$.2012-05-17

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HINT: If $\mathbf{x}$ satisfies $A \mathbf{x} = \mathbf{b}$, then $\dfrac{\mathbf{x}}{\alpha}$, where $\alpha$ is a scalar, satisfies $A \left(\dfrac{\mathbf{x}}{\alpha} \right) = \dfrac{\mathbf{b}}{\alpha}$. In your case, take $\alpha = d I$.

Equivalently, if you solve for $dc_1$, $dc_2$ and $d \lambda$, the solution will be of the form \begin{align} dc_1 & = f_1(u_{11},u_{12},u_{21},u_{22},p_1,p_2) dI\\ dc_2 & = f_2(u_{11},u_{12},u_{21},u_{22},p_1,p_2) dI\\ d \lambda & = f_3(u_{11},u_{12},u_{21},u_{22},p_1,p_2) dI \end{align} Hence, you will get \begin{align} \frac{dc_1}{dI} & = f_1(u_{11},u_{12},u_{21},u_{22},p_1,p_2)\\ \frac{dc_2}{dI} & = f_2(u_{11},u_{12},u_{21},u_{22},p_1,p_2)\\ \frac{d\lambda}{dI} & = f_3(u_{11},u_{12},u_{21},u_{22},p_1,p_2) \end{align}

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    I got you. Thanks you guys. Very helpful.2012-05-17