Answer to Question #160244 in Linear Algebra for Nikhil

Question

Obtain a solution set for the linear system

x1-2x2-3x3=0

-2x1+4x2+6x3=0

X1+2x2-5=0

Answer 1

Obtain a solution set for the linear system

x₁ - 2x₂ - 3x₃=0

-2x₁+4x₂+6x₃=0

X₁+2x₂-5x₃=0

$\begin{pmatrix} 1 & -2 &-3\\-2 &4 &6\\1 &2 &-5 \end{pmatrix}$ $\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\0 \end{pmatrix}$

$\begin{pmatrix} 1 & -2 &-3&| &0\\-2 &4 &6&|&0\\1 &2 &-5&|&0 \end{pmatrix}R_3 \mapsto2R_3 +R_2$ $\begin{pmatrix} 1 & -2 &-3&| &0\\-2 &4 &6&|&0\\0 &8 &-4&|&0 \end{pmatrix}$

$\begin{pmatrix} 1 & -2 &-3&| &0\\-2 &4 &6&|&0\\0 &8 &-4&|&0 \end{pmatrix} R_2 \mapsto R_2 +2R_1$ $\begin{pmatrix} 1 & -2 &-3&| &0\\0 &0 &0&|&0\\0 &8 &-4&|&0 \end{pmatrix}$

$x_1-2x_2-3x_3=0........(i)$

$8x_2-4x_3=0..........................(ii)$

from equation (ii)

$8x_2=4x_3 \implies x_3=2x_2$

we can view as determining x₃ in term of x₂ , with no restriction placed on x₂ a all

$x_2$ =$\alpha$ (arbitrary i.e. $\alpha$ can be any number)

then $x_3=2\alpha$

put x₂ and x₃ into eqn(i)

$x_1-2(\alpha)-3(2\alpha)=0$

$x_1=8\alpha$

$\because$ the linear system contains one free parameter