# Solving matrices for certain variables using Cramer's rule

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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Does it make a big difference? I have not taken linear algebra, so I do not know if placing matrices in different order changes something other than the conventional way of placing them according to cramer's rule. –  Koba May 17 '12 at 5:02
Great. First off, you try to compute the determinant of that coefficient matrix you have. That will be the denominator of your three solutions. You obtain the numerators of the three solutions by successively replacing columns of your coefficient matrix with the right-hand-side vector, and then taking the determinant of that. For instance, if you replace the first column, the determinant is the numerator of $dc_1$. –  J. M. is back. May 17 '12 at 5:04
Yes, it makes a big difference, since matrix multiplication is not commutative; you can't swap order. –  J. M. is back. May 17 '12 at 5:04
So something like that: $$\frac{\text{dc}_1}{\text{dI}}=\frac{\text{Det} \left( \begin{array}{ccc} 0 & u_{12} & -p_1 \\ 0 & u_{22} & -p_2 \\ \text{dI} & p_2 & 0 \end{array} \right)}{\text{Det} \left( \begin{array}{ccc} u_{11} & u_{12} & -p_1 \\ u_{21} & u_{22} & -p_2 \\ p_1 & p_2 & 0 \end{array} \right)}$$ –  Koba May 17 '12 at 5:21
@Dostre The solution you have written above is for $dc_1$ and not $$\dfrac{dc_1}{dI}$$ –  user17762 May 17 '12 at 5:24

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}