Question

find the basis and the dimension of the solution space W of the following system

X1 + 2X2 - 2X3 + 2X4 - X5 = 0

X1 + 2X2 - X3 + 3X4 - 2X5 = 0

2X1 + 4X2 - 7X3 +X4 - X5 = 0

Answer 1

Let us find the basis and the dimension of the solution space $W$ of the following system

$\begin{cases} X_1 + 2X_2 - 2X_3 + 2X_4 - X_5 = 0\\ X_1 + 2X_2 - X_3 + 3X_4 - 2X_5 = 0\\ 2X_1 + 4X_2 - 7X_3 +X_4 - X_5 = 0 \end{cases}$

After substracting from the second equation the first equation, and after multiplying the first equation by $-2$ and adding to the third equation, we get the following equivalent system:

$\begin{cases} X_1 + 2X_2 - 2X_3 + 2X_4 - X_5 = 0\\ X_3 + X_4 - X_5 = 0\\ -3X_3 -3X_4 +X_5 = 0 \end{cases}$

Let us multiply the second equation by 3 and add to the third equation:

$\begin{cases} X_1 + 2X_2 - 2X_3 + 2X_4 - X_5 = 0\\ X_3 + X_4 - X_5 = 0\\ -2X_5 = 0 \end{cases}$

It follows that

$\begin{cases} X_1 =- 2X_2 +2X_3 -2X_4 \\ X_3 =- X_4\\ X_5 = 0 \end{cases}$

and hence

$\begin{cases} X_1 =- 2X_2 -4X_4 \\ X_3 =- X_4\\ X_5 = 0 \end{cases}$

We conclude that

$W=\{(X_1,X_2,X_3,X_4,X_5)\in\R^5: X_1 =- 2X_2 -4X_4,\ X_3 =- X_4, \ X_5 = 0\}\\ =\{(- 2X_2 -4X_4,X_2,-X_4,X_4,0)\in\R^5: X_2, X_4 \in\R\}\\ =\{X_2(- 2,1,0,0,0)+X_4(-4,0,-1,1,0)\in\R^5: X_2, X_4 \in\R\}.$

Therefore, $\{(- 2,1,0,0,0),(-4,0,-1,1,0)\}$ is a basis for the solution space $W$ of the system, and hence $dim\ W=2.$