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

1. Solve the following equations by Cramer’s rule:

$\\ -x+y+z=a\\ x+y+z=b\\ x+y-z=c$

Solution:

Given: System of simultaneous equations

$\\ -x+y+z=a\\ x+y+z=b\\ x+y-z=c$

To find: The values of x, y and z

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

$\begin{bmatrix} -1\ \ \ \ \ \ \ \1\ \ \ \ \ \ \ 1\\ \ \1\ \ \ \ \ \ \1\ \ \ \ \ \ \ \1\\ \ \ \ \1\ \ \ \ \ \ \ \ \ 1\ \ \ \ -1\\ \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} a\\ b\\ c\\ \end{bmatrix}$

Let,

$A=\begin{bmatrix} -1\ \ \ \ \ \ \ \1\ \ \ \ \ \ \ 1\\ \ \1\ \ \ \ \ \ \1\ \ \ \ \ \ \ \1\\ \ \ \ \1\ \ \ \ \ \ \ \ \ 1\ \ \ \ -1\\ \end{bmatrix},\ X= \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}\ and\ d= \begin{bmatrix} a\\ b\\ c\\ \end{bmatrix}$

Now,

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

Next,

$A_{1}=\begin{bmatrix} a\ \ \ \ \ \ \ \1\ \ \ \ \ \ \ \1\\ b\ \ \ \ \ \ \ \1\ \ \ \ \ \ \ \1\\ \ c\ \ \ \ \ \ \ \ \ 1\ \ \ \ -1\\ \end{bmatrix}\ A_{2}=\begin{bmatrix} -1\ \ \ \ \ \ \ \ a\ \ \ \ \ \ \ \ 1\\ \ \ \1\ \ \ \ \ \ \ \ \ b\ \ \ \ \ \ \ \1\\ \ \ \1\ \ \ \ \ \ \ \ c\ \ \ \ -1\\ \end{bmatrix}\ A_{3}=\begin{bmatrix} -1\ \ \ \ \ \ \ \1\ \ \ \ \ \ \ a\\1\ \ \ \ \ \ \ \ \1\ \ \ \ \ \ \ \ b\\ 1\ \ \ \ \ \ \ \ \ \1\ \ \ \ \ \ \ \ c\\ \end{bmatrix}$

$\\\left |A_{1} \right |=a\begin{vmatrix} 1\ \ \ \ \1\\ 1\ -1 \end{vmatrix}-1 \begin{vmatrix} b \ \ \ \ \1\\ c \ -1 \end{vmatrix}+1 \begin{vmatrix} b\ \ \ 1 \\ c\ \ \ 1 \end{vmatrix}\\\\ \left |A_{1} \right |=a(-1-1)-1(-b-c)+1(b-c)\\ \left |A_{1} \right |=a(-2)+b+c+b-c\\ \left |A_{1} \right |=-2a+2b\\ \left |A_{1} \right |=-2(a-b)\\$

$\\\left |A_{2} \right |=-1\begin{vmatrix} b\ \ \ \ \1\\ c\ -1 \end{vmatrix}-a \begin{vmatrix} 1 \ \ \ \ \1\\ 1 \ -1 \end{vmatrix}+1 \begin{vmatrix} 1\ \ \ b \\ 1\ \ \ c \end{vmatrix}\\\\ \left |A_{2} \right |=-1(-b-c)-a(-1-1)+1(c-b)\\ \left |A_{2} \right |=b+c-a(-2)+c-b\\ \left |A_{2} \right |=2c+2a\\ \left |A_{2} \right |=2a+2c\\ \left |A_{2} \right |=2(a+c)\\$

$\\\left |A_{3} \right |=-1\begin{vmatrix} 1\ \ \ b\\ 1\ \ \ c \end{vmatrix}-1 \begin{vmatrix} 1 \ \ \ b\\ 1 \ \ \ c \end{vmatrix}+a \begin{vmatrix} 1\ \ \1 \\ 1\ \ \1 \end{vmatrix}\\\\ \left |A_{3} \right |=-1(c-b)-1(c-b)+a(1-1)\\ \left |A_{3} \right |=-c+b-c+b+a(0)\\ \left |A_{3} \right |=-2c+2b+0\\ \left |A_{3} \right |=2b-2c\\ \left |A_{3} \right |=2(b-c)\\$

$\\\therefore x=\frac{\left |A_{1}\right |}{\left |A\right |}=\frac{-2(a-b)}{4}=-\frac{1}{2}(a-b)\\ y=\frac{\left |A_{2}\right |}{\left |A\right |}=\frac{2(a+c)}{4}=\frac{1}{2}(a+c)\\ z=\frac{\left |A_{3}\right |}{\left |A\right |}=\frac{2(b-c)}{4}=\frac{1}{2}(b-c)\\$

$\therefore x=-\frac{1}{2}(a-b),\ y=\frac{1}{2}(a+c),\ and\ z=\frac{1}{2}(b-c)\\$

Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Cramer's Rule - Solution to System of Simultaneous Equations

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