Answer to Question #123418 in Linear Algebra for zwonaka Dzivhani

Question

On a farmer’s market in Essex, the demand and supply functions for (a kilogram (kg) of) oranges (1), (a kg of) lemon (2) and (a kg of) nectarines (3) are:
q_1^d=-2p_1-4p_2+2p_3+160,q_1^s=2p_1+80
q_2^d=-4p_1-2p_2+5p_3+200,q_2^s=3p_2+160
q_3^d=2p_1+4p_2-5p_3+300,q_3^s=p_3+90
Use the equilibrium condition q_i^d=q_i^s to rewrite this system in the form Ap=d, where A is a 3×3 matrix of coefficients, p=(■(p_1@p_2@p_3 )), and d is a 3×1 vector. Then, use Cramer’s rule to solve for p_1 and p_3 (ONLY solve for p_1 and p_3!)

Answer 1

$$-2p_1-4p_2+2p_3+160=2p_1+80$$ $$-4p_1-2p_2+5p_3+200=3p_2+160$$ $$2p_1+4p_2-5p_3+300=p_3+90$$

$$-4p_1-4p_2+2p_3=-80$$ $$-4p_1-5p_2+5p_3=-40$$ $$2p_1+4p_2-6p_3=-210$$

$$A=\begin{pmatrix} -4 & -4 & 2 \\ -4 & -5 & 5 \\ 2 & 4 & -6 \end{pmatrix}$$

$$p=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}, d=\begin{pmatrix} -80 \\ -40 \\ -210 \end{pmatrix}$$

$$Ap=d$$

$$\Delta=\begin{vmatrix} -4 & -4 & 2 \\ -4 & -5 & 5 \\ 2 & 4 & -6 \end{vmatrix}=$$

$$=-4\begin{vmatrix} -5 & 5 \\ 4 & -6 \end{vmatrix}-(-4)\begin{vmatrix} -4 & 5 \\ 2 & -6 \end{vmatrix}+2\begin{vmatrix} -4 & -5 \\ 2 & 4 \end{vmatrix}=$$

$$=-4(30-20)+4(24-10)+2(-16+10)=4\not=0$$

Solutions exist.

$$\Delta_1=\begin{vmatrix} -80 & -4 & 2 \\ -40 & -5 & 5 \\ -210 & 4 & -6 \end{vmatrix}=$$

$$=-80\begin{vmatrix} -5 & 5 \\ 4 & -6 \end{vmatrix}-(-4)\begin{vmatrix} -40 & 5 \\ -210 & -6 \end{vmatrix}+2\begin{vmatrix} -40 & -5 \\ -210 & 4 \end{vmatrix}=$$

$$=-80(30-20)+4(240+1050)[phone])=1940$$

$$\Delta_3=\begin{vmatrix} -4 & -4 & -80 \\ -4 & -5 & -40 \\ 2 & 4 & -210 \end{vmatrix}=$$

$$=-4\begin{vmatrix} -5 & -40 \\ 4 & -210 \end{vmatrix}-(-4)\begin{vmatrix} -4 & -40 \\ 2 & -210 \end{vmatrix}-80\begin{vmatrix} -4 & -5 \\ 2 & 4 \end{vmatrix}=$$

$$=-4(1050+160)+4(840+80)-80(-16+10)=-680$$

$$p_1={1940\over 4}=485$$

$$p_3={-680\over 4}=-170$$