Answer to Question #179236 in Linear Algebra for Aditya Sharma

Question

Consider the system of equations

11 2x1 + x2 + 4x3 =

3x1 + x2 + 5x3 =14

A feasible solution is x1 = ,2 x2 = ,3 x3 = .1 Reduce this feasible solution to a basic

feasible solution.

Answer 1

$\begin{cases} 2x_1+x_2=11-4x_3, \\ 3x_1+x_2=14-5x_3; \end{cases}$

$\Delta=\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix}=2-3=-1,$

$\Delta_1=\begin{vmatrix} 11-4x_3& 1 \\ 14-5x_3& 1 \end{vmatrix}=11-4x_3-14+5x_3=x_3-3,$

$\Delta_2=\begin{vmatrix} 2 & 11-4x_3 \\ 3 & 14-5x_3 \end{vmatrix}=28-10x_3-33+12x_3=2x_3-5,$

$x_1=\frac{\Delta_1}{\Delta}=3-x_3,$

$x_2=\frac{\Delta_2}{\Delta}=5-2x_3.$