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Problem 8: Solve the following system by (a) Gaussian elimination method (b) Gauss-Jordan elimination method,

 Problem 8: Solve the following system by a. Gaussian elimination method b. Gauss-Jordan elimination method, c. Find particular solution of the system $AX = B$.

\begin{align*} x_1+2x_2+x_3+x_4=1\\ 3x_1+4x_4=1\\ x_1-4x_2-2x_3-2x_4=0 \end{align*}

Solution:

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

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrrr|r} 1 & 2 & 1 & 1 & 1 \\ 3 & 0& 0 & 4 & 1 \\ 1 & -4 & -2 & -2 & 0 \\ \end{array}\right] \xrightarrow{R_2\rightarrow R_2-3R_1} \left[\begin{array}{rrrr|r} 1 & 2 & 1 & 1 & 1 \\ 0 & -6& -3 & 1 & -2 \\ 1 & -4 & -2 & -2 & 0 \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R3\rightarrow R_3-R_1} \left[\begin{array}{rrrr|r} 1 & 2 & 1 & 1 & 1 \\ 0 & -6& -3 & 1 & -2 \\ 0 & -6 & -3 & -1 & -1 \\ \end{array}\right] \xrightarrow{R3\rightarrow R_3-R_2} \left[\begin{array}{rrrr|r} 1 & 2 & 1 & 1 & 1 \\ 0 & -6& -3 & 1 & -2 \\ 0 & 0 & 0 & -2 & 1 \\ \end{array}\right]\\ \xrightarrow[R2\rightarrow \frac{-1}{6}R_2]{R3\rightarrow \frac{-1}{2}R_3} \left[\begin{array}{rrrr|r} 1 & 2 & 1 & 1 & 1 \\ 0 & 1& \frac{1}{2} & \frac{-1}{6} & \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{-1}{2} \\ \end{array}\right] ...(REF)\\ \xrightarrow{R1\rightarrow R_1-R_2} \left[\begin{array}{rrrr|r} 1 & 0 & 0 & \frac{4}{3} & \frac{1}{3} \\ 0 & 1& \frac{1}{2} & \frac{-1}{6} & \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{-1}{2} \\ \end{array}\right]\\ \xrightarrow[R2\rightarrow R2+\frac{1}{6}R_3]{R1\rightarrow R1-\frac{4}{3}R_3} \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1 \\ 0 & 1& \frac{1}{2} & 0 & \frac{1}{4} \\ 0 & 0 & 0 & 1 & \frac{-1}{2} \\ \end{array}\right] ...(RREF)\]

(a). For Gaussian elimination we write down the equations corresponding to REF. Then we get

\begin{align*} x_1+2x_2+x_3+x_4=1\\ x_2+\frac{1}{2}x_3-\frac{-1}{6}x_4=\frac{1}{3}\\ x_4=\frac{-1}{2} \end{align*}

Expresing $x_1$ and $x_2$ we get

\begin{align*} x_1 =\frac{1}{3}-\frac{4}{3}x_4\\ x_2 =\frac{1}{3}-\frac{1}{2}x_3+\frac{1}{6}x_4 \end{align*}

Substitute $x_4 = -\frac{1}{2}$, Solution set S is \begin{align*} x_1=1\\ x_2=\frac{1}{4}-\frac{1}{2}t\\ x_3=t\\ x_4 = -\frac{1}{2}, t\in \mathbb{R} \end{align*}

\begin{align*} S=\left\{1,\frac{1}{4}-\frac{1}{2}t,t,-\frac{1}{2}|t\in \mathbb{R}\right\}\\ =\left\{(1,\frac{1}{4},0,-\frac{1}{2})+t(0,-\frac{1}{2},1,0)|t\in \mathbb{R}\right\} \end{align*}

(b). For Gauss-Jordan elimination we write down the equations corresponding to RREF. Then we get

\begin{align*} x_1=1\\ x_2+\frac{1}{2}x_3=\frac{1}{4}\\ x_4 = -\frac{1}{2} \end{align*}

Solution set S is

\begin{align*} S=\left\{1,\frac{1}{4}-\frac{1}{2}t,t,-\frac{1}{2}|t\in \mathbb{R}\right\}\\ =\left\{(1,\frac{1}{4},0,-\frac{1}{2})+t(0,-\frac{1}{2},1,0)|t\in \mathbb{R}\right\} \end{align*}

(c). $(1,\frac{1}{4},0,-\frac{1}{2})$ is a particular solution of $AX=B$ and $t(0,-\frac{1}{2},1,0)$ is arbitrary solution of $AX=0$.

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