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

1. Solve the following equations by Matrix Inverse method and Cramer's Rule:

$\\\8x_{1}-x_{2}=16\\ 2x_{2}+5x_{3}=5\\ 2x_{1}+3x_{3}=7\\$

Solution:

Given: System of Simultaneous equations

$\\\8x_{1}-x_{2}=16\\ 2x_{2}+5x_{3}=5\\ 2x_{1}+3x_{3}=7\\$

or,

$\\\8x_{1}-x_{2}+0.x_{3}=16\\ 0.x_{1}+2x_{2}+5x_{3}=5\\ 2x_{1}+0.x_{2}+3x_{3}=7\\$

To find: $x_{1},\ x_{2}\ and\ x_{3}$

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

$\begin{bmatrix} 8\ -1\ \ \ \ \ \0\\ 0\ \ \ \ \2 \ \ \ \ \ \5\\ 2\ \ \ \ \ 0\ \ \ \ \ 3 \\ \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}= \begin{bmatrix} 16\\ 5\\ 7 \end{bmatrix}$

Let,

$A=\begin{bmatrix} 8\ -1\ \ \ \ \ \0\\ 0\ \ \ \ \2 \ \ \ \ \ \5\\ 2\ \ \ \ \ 0\ \ \ \ \ 3 \\ \end{bmatrix},\ X= \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}\ and\ \ d= \begin{bmatrix} 16\\ 5\\ 7 \end{bmatrix}$

Now,

$\\\left |A \right |=8\begin{vmatrix} 2\ \ 5\\ 0\ \ 3 \end{vmatrix}-(-1) \begin{vmatrix} 0\ \5\\ 2\ \ 3 \end{vmatrix}+0 \begin{vmatrix} 0\ \2\\ 2\ \0 \end{vmatrix}\\\\ \left |A \right |=8(6-0)+1(0-10)+0(0-4)\\ \left |A \right |=8(6)+1(-10)+0(-4)\\ \left |A \right |=48-10+0\\ \left |A \right |=48-10\\ \left |A \right |=38\\$

Next,

$A_{1}=\begin{bmatrix} 16\ -1\ \ \ \0\\ 5\ \ \ \ \2\ \ \ \ \5\\ 7\ \ \ \ \0\ \ \ \ \3 \end{bmatrix}\ A_{2}= \begin{bmatrix} 8\ \ \ \16\ \ \ \0\\ 0\ \ \ \ \5\ \ \ \ \5\\ 2\ \ \ \ \7\ \ \ \ \3 \end{bmatrix}\ A_{3}= \begin{bmatrix} 8\ \ -1\ \ \ \16\\ 0\ \ \ \ \ 2\ \ \ \ \5\\ 2\ \ \ \ \0\ \ \ \ \7 \end{bmatrix}$

$\\\left |A_{1} \right |=16\begin{vmatrix} 2\ \ 5\\ 0\ \ 3 \end{vmatrix}-(-1) \begin{vmatrix} 5\ \ 5\\ 7\ \ 3 \end{vmatrix}+0 \begin{vmatrix} 5\ \ 2\\ 7\ \ 0 \end{vmatrix}\\\\ \left |A_{1} \right |=16(6-0)+1(15-35)+0(0-14)\\ \left |A_{1} \right |=16(6)+1(-20)+0(-14)\\ \left |A_{1} \right |=96-20+0\\ \left |A_{1} \right |=96-20\\ \left |A_{1} \right |=76$

$\\\left |A_{2} \right |=8\begin{vmatrix} 5\ \ 5\\ 7\ \ 3 \end{vmatrix}-16 \begin{vmatrix} 0\ \ 5\\ 2\ \ 3 \end{vmatrix}+0 \begin{vmatrix} 0\ \ 5\\ 2\ \ 7 \end{vmatrix}\\\\ \left |A_{2} \right |=8(15-35)-16(0-10)+0(0-10)\\ \left |A_{2} \right |=8(-20)-16(-10)+0(-10)\\ \left |A_{2} \right |=-160+160+0\\ \left |A_{2} \right |=-160+160\\ \left |A_{2} \right |=0$

$\\\left |A_{3} \right |=8\begin{vmatrix} 2\ \ 5\\ 0\ \ 7 \end{vmatrix}-(-1) \begin{vmatrix} 0\ \ 5\\ 2\ \ 7 \end{vmatrix}+16 \begin{vmatrix} 0\ \ 2\\ 2\ \ 0 \end{vmatrix}\\\\ \left |A_{3} \right |=8(14-0)+1(0-10)+16(0-4)\\ \left |A_{3} \right |=8(14)+1(-10)+16(-4)\\ \left |A_{3} \right |=112-10-64\\ \left |A_{3} \right |=112-74\\ \left |A_{3} \right |=38$

$\\\therefore x=\frac{\left |A_{1}\right |}{\left |A\right |}=\frac{76}{38}=2\\ y=\frac{\left |A_{2}\right |}{\left |A\right |}=\frac{0}{38}=0\\ z=\frac{\left |A_{3}\right |}{\left |A\right |}=\frac{38}{38}=1\\$

$\\\therefore x=2\ ,\ y=0\ and\ z=1\\$ .

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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