Economics at Zero Marginal Cost: How to solve Simultaneous Equations by Cramer’s Rule (3 equations & 3 Variables)

1. The equilibrium condition for three markets is given by

$11P_{1}-P_{2}-P_{3}=31\\ -P_{1}+6P_{2}-2P_{3}=26\\ -P_{1}-2P_{2}+7P_{3}=24$

Using Cramer's Rule, find the equilibrium price for each market.

Solution:

Given: System of simultaneous equations

$11P_{1}-P_{2}-P_{3}=31\\ -P_{1}+6P_{2}-2P_{3}=26\\ -P_{1}-2P_{2}+7P_{3}=24$

To find: Equilibrium values for each market, i.e., $P_{1},\ P_{2},\ and\ P_{3}$

Re-writing the set of equations in matrix form:

$\begin{bmatrix} 11\ \ \ -1\ \ \ -1\\ -1\ \ \ \ \ \ 6\ \ -2\\ -1\ \ \ -2\ \ \ \ \ \ 7\\ \end{bmatrix} \begin{bmatrix} P_{1}\\ P_{2}\\ P_{3}\\ \end{bmatrix}= \begin{bmatrix} 31\\ 26\\ 24\\ \end{bmatrix}$

Let

$A=\begin{bmatrix} 11\ \ \ -1\ \ \ -1\\ -1\ \ \ \ \ \ 6\ \ -2\\ -1\ \ \ -2\ \ \ \ \ \ 7\\ \end{bmatrix},\ X=\begin{bmatrix} P_{1}\\ P_{2}\\ P_{3}\\ \end{bmatrix}\ and\ d= \begin{bmatrix} 31\\ 26\\ 24\\ \end{bmatrix}$

Now,

$\\\left |A \right |=11\begin{vmatrix} 6\ -2 \\ -2\ \ \ 7 \end{vmatrix}-(-1) \begin{vmatrix} -1\ -2 \\ -1\ \ \ \7 \end{vmatrix}+(-1) \begin{vmatrix} -1\ \ \ \ \6 \\ -1\ \ -2 \end{vmatrix}\\\\ \left |A \right |=11(42-4)+1(-7-2)-1(2-(-6))\\ \left |A \right |=11(38)+1(-9)-1(2+6)\\ \left |A \right |=418-9-1(8)\\ \left |A \right |=418-9-8\\ \left |A \right |=418-17\\ \left |A \right |=401$

Next,

$A_{1}=\begin{bmatrix} 31\ -1\ -1\\ 26\ \ \ \ 6\ \ -2 \\ 24\ -2\ \ \ \ \ 7 \end{bmatrix}\ \ A_{2}=\begin{bmatrix} 11\ \ \ \31\ \ -1\\ -1\ \ \ 26\ \ -2 \\ -1 \ \ \ 24\ \ \ \ \ \ 7 \end{bmatrix}\ \ A_{3}=\begin{bmatrix} 11\ \ \ -1\ \ \ \ \ \ 31\\ -1 \ \ \ \ \6\ \ \ \ \ \ 26 \\ -1\ \ -2\ \ \ \ \ 24 \end{bmatrix}$

$\\\left |A_{1} \right |=31\begin{vmatrix} 6\ -2 \\ -2\ \ \ 7 \end{vmatrix}-(-1) \begin{vmatrix} 26\ -2 \\ 24\ \ \ \7 \end{vmatrix}+(-1) \begin{vmatrix} 26\ \ \ \ \6 \\ 24\ \ -2 \end{vmatrix}\\\\ \left |A_{1} \right |=31(42-4)+1(182-(-48))-1(-52-144)\\ \left |A_{1} \right |=31(38)+1(182+48)-1(-196)\\ \left |A_{1} \right |=1178+1(230)+196\\ \left |A_{1} \right |=1178+230+196\\ \left |A_{1} \right |=1604$

$\\\left |A_{2} \right |=11\begin{vmatrix} 26\ -2 \\ 24\ \ \ \ \ 7 \end{vmatrix}-31 \begin{vmatrix} -1\ -2 \\ -1\ \ \ \7 \end{vmatrix}+(-1) \begin{vmatrix} -1\ \ \ \ \26 \\ -1\ \ \ \ \24 \end{vmatrix}\\\\ \left |A_{2} \right |=11(182-(-48))-31(-7-2)-1(-24-(-26))\\ \left |A_{2} \right |=11(182+48)-31(-9)-1(-24+26)\\ \left |A_{2} \right |=11(230)+279-1(2)\\ \left |A_{2} \right |=2530+279-2\\ \left |A_{2} \right |=2807$

$\\\left |A_{3} \right |=11\begin{vmatrix} 6\ \ \ \ \ \ \26 \\ -2\ \ \ \ \ 24 \end{vmatrix}-(-1) \begin{vmatrix} -1\ \ \ \26\\ -1\ \ \ \24 \end{vmatrix}+31 \begin{vmatrix} -1\ \ \ \ \ \6 \\ -1\ \ \ -2 \end{vmatrix}\\\\ \left |A_{3} \right |=11(144-(-52))+1(-24-(-26))+31(2-(-6))\\ \left |A_{3} \right |=11(144+52)+1(-24+26)+31(2+6)\\ \left |A_{3} \right |=11(196)+1(2)+31(8)\\ \left |A_{3} \right |=2156+2+248\\ \left |A_{3} \right |=2406$

$\\\therefore P_{1}=\frac{\left |A_{1}\right |}{\left |A\right |}=\frac{1604}{401}=4\\ P_{2}=\frac{\left |A_{2}\right |}{\left |A\right |}=\frac{2807}{401}=7\\ P_{3}=\frac{\left |A_{3}\right |}{\left |A\right |}=\frac{2406}{401}=6\\$