Question

Obtain all the basic solutions of the following linear equations:
2x1 + x2 + x3 + x4 = 6
3x1 + 2x2 + x3 + 2x2 = 8

Answer 1

$Since \ we \ have \\ 2x_1 + x_2 + x_3 + x_4 = 6\\ 3x_1 + 2x_2 + x_3 + 2x_4 = 8\\ This \ system \ can \ written\ in \ the \ form\\ \begin{bmatrix} 2 & 1&1&1\\ 3&2&1&2 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix}=\begin{bmatrix} 6\\ 8 \end{bmatrix}\\ \text{Using Gauss elimination, we get }\\ \begin{bmatrix} \begin{array}{cccc|c} 2 & 1 &1&1&6 \\ 3& 2 & 1&2&8 \\ \end{array} \end{bmatrix} \xrightarrow {R_2-R_1\rightarrow R_2}\\ \\ \\ \begin{bmatrix} \begin{array}{cccc|c} 2 & 1 &1&1&6 \\ 0& -1 &0&1&2 \\ \end{array} \end{bmatrix}\xrightarrow {-R_2\rightarrow R_2}\\ \\ \\ \begin{bmatrix} \begin{array}{cccc|c} 3 & 1 &1&1&6 \\ 0& 1 &0&-1&-2 \\ \end{array} \end{bmatrix}\\ \text{So, we get the rank of matrix is 2}\\ \text{ and the number of unknowns is 4}\\ \text {this implies , we have 2 free unknown}\\ \text{So, let}\\ x_3=t , x_4=s.\\ \text{Now, we have}\\ x_2-x_4=4\Rightarrow x_2=4-s\\ 2x_1 + x_2 + x_3 + x_4 = 6\\ \Rightarrow 2x_1 =- x_2 - t -s+ 6\\ 2x_1 =-4+s - t -s+ 6\\ 2x_1 =2-t\\$