Economics at Zero Marginal Cost: Cramer's Rule - Solution to System of Simultaneous Equations

1. Solve the following system of equations using Cramer’s rule:

$\\\ -x+3z+1=0\\ 2x-y-4z-2=0\\ y+2z-4=0$

Solution:

Given: System of simultaneous equations

$\\\ -x+3z+1=0\\ 2x-y-4z-2=0\\ y+2z-4=0$

or,

$\\\ -x+3z=-1\\ 2x-y-4z=2\\ y+2z=4$

or,

$\\\ -x+0.y+3z=-1\\ 2x-y-4z=2\\ 0.x+y+2z=4$

Re-writing the above set of equations in matrix forms:

$\begin{bmatrix} -1\ \ \ \ 0\ \ \ \3 \\ 2\ \ -1\ -4\\ 0\ \ \ \ \ \1\ \ \ \ \2 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix}= \begin{bmatrix} -1\\ 2\\ 4 \end{bmatrix}$

Let,

$A=\begin{bmatrix} -1\ \ \ \ 0\ \ \ \3 \\ 2\ \ -1\ -4\\ 0\ \ \ \ \ \1\ \ \ \ \2 \end{bmatrix},\ X= \begin{bmatrix} x\\ y\\ z \end{bmatrix}\ and\ d= \begin{bmatrix} -1\\ 2\\ 4 \end{bmatrix}$

$\\\left |A \right |=-1\begin{vmatrix} -1\ -4 \\ 1\ \ \ \ \2 \end{vmatrix}-0 \begin{vmatrix} 2\ \ -4 \\ 0\ \ \ \ \2 \end{vmatrix}+3 \begin{vmatrix} 2\ \ \ \ -1 \\ 0\ \ \ \ \ \1 \end{vmatrix}\\\\ \left |A \right |=-1(-2-(-4))-0(4-0)+3(2-0)\\ \left |A \right |=-1(-2+4)-0(4)+3(2)\\ \left |A \right |=-1(2)-0+6\\ \left |A \right |=-2+6\\ \left |A \right |=4\\$

Next,

$A_{1}=\begin{bmatrix} -1\ \ \ \ 0\ \ \ \3 \\ 2\ \ -1\ -4\\ 4\ \ \ \ \ \1\ \ \ \ \2 \end{bmatrix}\ A_{2}=\begin{bmatrix} -1\ \ \ \ -1\ \ \ \3 \\ 2\ \ \ \ \ \ \ 2\ \ \ -4\\ 0\ \ \ \ \ \4\ \ \ \ \ \ \2 \end{bmatrix}\ A_{3}=\begin{bmatrix} -1\ \ \ \ 0\ \ \ \ -1 \\ 2\ \ \ -1\ \ \ \ \ \2\\ 0\ \ \ \ \ \1\ \ \ \ \ \ \4 \end{bmatrix}$

$\\\left |A_{1} \right |=-1\begin{vmatrix} -1\ -4 \\ 1\ \ \ \ \2 \end{vmatrix}-0 \begin{vmatrix} 2\ \ -4 \\ 4\ \ \ \ \2 \end{vmatrix}+3 \begin{vmatrix} 2\ \ \ \ -1 \\ 4\ \ \ \ \ \1 \end{vmatrix}\\\\ \left |A_{1} \right |=-1(-2-(-4))-0(4-(-16))+3(2-(-4))\\ \left |A_{1} \right |=-1(-2+4)-0(4+16)+3(2+4)\\ \left |A_{1} \right |=-1(2)-0(20)+3(6)\\ \left |A_{1} \right |=-2-0+18\\ \left |A_{1} \right |=-2+18\\ \left |A_{1} \right |=16\\$

$\\\left |A_{2} \right |=-1\begin{vmatrix} 2\ -4 \\ 4\ \ \ \ \2 \end{vmatrix}-(-1) \begin{vmatrix} 2\ \ -4 \\ 0\ \ \ \ \2 \end{vmatrix}+3 \begin{vmatrix} 2\ \ \ \ \2 \\ 0\ \ \ \ \ \4 \end{vmatrix}\\\\ \left |A_{2} \right |=-1(4-(-16))+1(4-0)+3(8-0)\\ \left |A_{2} \right |=-1(4+16)+1(4)+3(8)\\ \left |A_{2} \right |=-1(20)+4+24\\ \left |A_{2} \right |=-20+4+24\\ \left |A_{2} \right |=-20+28\\ \left |A_{2} \right |=8\\$

$\\\left |A_{3} \right |=-1\begin{vmatrix} -1\ \ \2 \\ 1\ \ \ \ \4 \end{vmatrix}-0 \begin{vmatrix} 2\ \ \2 \\ 0\ \ \4 \end{vmatrix}+(-1) \begin{vmatrix} 2\ \ \ -1 \\ 0\ \ \ \ \ \1 \end{vmatrix}\\\\ \left |A_{3} \right |=-1(-4-2)-0(8-0)-1(2-0)\\ \left |A_{3} \right |=-1(-6)-0(8)-1(2)\\ \left |A_{3} \right |=6-0-2\\ \left |A_{3} \right |=6-2\\ \left |A_{3} \right |=4\\$

$\\\therefore x=\frac{\left |A_{1}\right |}{\left |A\right |}=\frac{16}{4}=4\\ y=\frac{\left |A_{2}\right |}{\left |A\right |}=\frac{8}{4}=2\\ z=\frac{\left |A_{3}\right |}{\left |A\right |}=\frac{4}{4}=1\\$