Economics at Zero Marginal Cost: How to solve Simultaneous Equations by Matrix Inversion (3 equations & 3 Variables)

1. Solve the following system of simultaneous equations by matrix inversion

$\\\4x_1+x_2-5x_3=8\\ -2x_1+3x_2+x_3=12\\ 3x_1-x_2+4x_3=5\\$

Solution:

Given: System of Simultaneous Equations

$\\\4x_1+x_2-5x_3=8\\ -2x_1+3x_2+x_3=12\\ 3x_1-x_2+4x_3=5\\$

Re-writing the equations in matrix form, we have

$\\\begin{bmatrix} 4\ \ \ 1\ -5\\ -2\ \ \ \ 3\ \ 1\\ 3\ -1\ \ 4\\ \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} = \begin{bmatrix} 8\\ 12\\ 5 \end{bmatrix}$

Let,

$\\A= \begin{bmatrix} 4\ \ \ 1\ -5\\ -2\ \ \ \ 3\ \ 1\\ 3\ -1\ \ 4\\ \end{bmatrix}, X=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} and\ d= \begin{bmatrix} 8\\ 12\\ 5 \end{bmatrix}$

We know that, $X=A^{-1}d$

Now,

$A^{-1}=\frac{1}{\left | A \right |}.Adj\ A$

$\\\left | A \right |=4\begin{vmatrix} 3\ \ \ 1 \\ -1\ 4 \end{vmatrix}-1\begin{vmatrix} -2\ \ 1 \\ 3\ \ \ 4 \end{vmatrix}+(-5)\begin{vmatrix} -2\ \ \ 3 \\ 3\ \ -1 \end{vmatrix}\\\ \\\left | A \right |=4(12-(-1))-1(-8-3)-5(2-9) \\\left | A \right |=4(12+1)-1(-11)-5(-7) \\\left | A \right |=4(13)+11+35 \\\left | A \right |=52+11+35 \\\left | A \right |=98$

$\because$ the value of ${\left | A \right |}\neq 0$

$\therefore$ matrix A is non-singular and $A^{-1}$ exist.

Next,

$\\Co-factor\ of\ a_{11},C_{11}\ i.e.,\ 4=\begin{vmatrix} 3\ \ \ 1 \\ -1\ 4 \end{vmatrix}=12-(-1)=12+1=13\\ Co-factor\ of\ a_{12},C_{12}\ i.e.,\ 1=-\begin{vmatrix} -2\ \ \ 1 \\ 3\ \ \ \ \ 4 \end{vmatrix}=-(-8-3)=-(-11)=11\\ Co-factor\ of\ a_{13},C_{13}\ i.e.,\ -5=\begin{vmatrix} -2\ \ \ 3 \\ 3\ \ -1 \end{vmatrix}=2-9=-7\\ Co-factor\ of\ a_{21},C_{21}\ i.e.,\ -2=-\begin{vmatrix} 1\ \ \ -5 \\ -1\ \ \ \ 4 \end{vmatrix}=-(4-5)=-(-1)=1\\ Co-factor\ of\ a_{22},C_{22}\ i.e.,\ 3=\begin{vmatrix} 4\ -5 \\ 3\ \ \ 4 \end{vmatrix}=16-(-15)=16+15=31\\ Co-factor\ of\ a_{23},C_{23}\ i.e.,\ 1=-\begin{vmatrix} 4\ \ \ \ 1 \\ 3\ -1 \end{vmatrix}=-(-4-3)=-(-7)=7\\ \\Co-factor\ of\ a_{31},C_{31}\ i.e.,\ 3=-\begin{vmatrix} 1\ -5 \\ 3\ \ \ \ 1 \end{vmatrix}=1-(-15)=1+15=16\\ Co-factor\ of\ a_{32},C_{32}\ i.e.,\ -1=-\begin{vmatrix} 4\ \ -5 \\ -2\ \ \ 1 \end{vmatrix}=-(4-10)=-(-6)=6\\ Co-factor\ of\ a_{33},C_{33}\ i.e.,\ 4=-\begin{vmatrix} 4\ \ \ \ 1 \\ -2\ \ 3 \end{vmatrix}=12-(-2)=1+2=14\\$

$\\Co-factor\ matrix\ of\ A = \begin{bmatrix} 13\ \ \ 11\ -7\\ 1\ \ \ \ 31\ \ \ \ \ 7\\ 16\ \ \ \ 6\ \ \ \ 14\\ \end{bmatrix}$

$\\Adjoint\ matrix\ of\ A,\ Adj\ A = \begin{bmatrix} 13\ \ \ \ 1\ \ \ \ \ 16\\ 11\ \ \ \ 31\ \ \ \ \ 6\\ -7\ \ \ \ 7\ \ \ \ 14\\ \end{bmatrix}$

Now,

$A^{-1}=\frac{1}{\left | A \right |}.Adj\ A$

$\\A^{-1} = \frac{1}{98}\begin{bmatrix} 13\ \ \ \ 1\ \ \ \ \ 16\\ 11\ \ \ \ 31\ \ \ \ \ 6\\ -7\ \ \ \ 7\ \ \ \ 14\\ \end{bmatrix}$

Again, $X=A^{-1}d$

$\\X= \frac{1}{98}\begin{bmatrix} 13\ \ \ \ 1\ \ \ \ \ 16\\ 11\ \ \ \ 31\ \ \ \ \ 6\\ -7\ \ \ \ 7\ \ \ \ 14\\ \end{bmatrix}\begin{bmatrix} 8\\ 12\\ 5 \end{bmatrix}$

$\\X = \frac{1}{98}\begin{bmatrix} 13\times8 \ +\ \ \ 1\times12\ \ + \ \ \ 16\times5\\ 11\times8\ +\ \ \ 31\times12\ +\ \ \ \ 6\times5\\ -7\times8 +\ \ \ 7\times12 \ \ \ +\ \ 14\times5\\ \end{bmatrix}$