Problem 7: Solve the following system of linear equations using Gaussian elimination.

 Problem 7: Solve the following system of linear equations using Gaussian elimination. \begin{align*} x_1+2x_2+x_3=5\\ x_1+x_2+5x_3+2x_4=7\\ x_1+2x_2+8x_3+4x_4=12 \end{align*}

Solution:

Consider the augmented matrix of this system and apply row operations.

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrrr|r} 1 & 0 & 2 & 1 & 5 \\ 1 & 1 & 5 & 2 & 7 \\ 1 & 2 & 8 & 4 & 12 \\ \end{array}\right] \xrightarrow{R_2\rightarrow R_2-R_1} \left[\begin{array}{rrrr|r} 1 & 0 & 2 & 1 & 5 \\ 0 & 1 & 3 & 1 & 2 \\ 1 & 2 & 8 & 4 & 12 \\ \end{array}\right]. \end{align*}\] \[\xrightarrow{R3\rightarrow R_3-R_1} \left[\begin{array}{rrrr|r} 1 & 0 & 2 & 1 & 5 \\ 0 & 1 & 3 & 1 & 2 \\ 0 & 2 & 6 & 3 & 7 \\ \end{array}\right] \xrightarrow{R3\rightarrow R_3-2R_2} \left[\begin{array}{rrrr|r} 1 & 0 & 2 & 1 & 5 \\ 0 & 1 & 3 & 1 & 2 \\ 0 & 0 & 0 & 1 & 3 \\ \end{array}\right]\]

For Gauss-elimination we write down the equations corresponding to row reduction form. Then we get

\begin{align*} x_1+2x_3+x_4=5\\ x_2+3x_3+x_4=2\\ x_4=3 \end{align*}

Using $x_4=3$ in first two equations and expresing $x_1$ and $x_2$ in terms of constants and $x_3$, we get

\begin{align*} x_1 =5-3-2x_3=2-2x_3\\ x_2=2-3-3x_3=-1-3x_3 \end{align*}

let $x_3$ = t, Solution set S is \begin{align*} x_1=2-2t\\ x_2=-1-3t\\ x_3=t\\ x_4 = 3, t\in \mathbb{R} \end{align*}

\begin{align*} S=\left\{2-2t,-1-3t,t,3|t\in \mathbb{R}\right\}\\ =\left\{(2,-1,0,3)+t(-2,-3,1,0)|t\in \mathbb{R}\right\} \end{align*}