I have a matrix which after doing row reduced echelon form is as follows:$$A=\begin{bmatrix}1&1&-1&-1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}.$$ When I try to solve $Ax = 0$, I am confused. $x_2,x_3,x_4$ are free variables because they have no pivot and only $x_1$ is pivoted variable. Following is the equation I get: $$x_1+x_2-x_3-x_4 = 0$$ What is the best way to find all the solutions to $Ax=0$?

  • $\begingroup$ Hyperplan equation ! $\endgroup$
    – Damien
    Dec 2 '18 at 9:42
  • $\begingroup$ In case the real issue is the X-Y Problem, please add the original matrix and your steps in row reduction, so we can verify your matrix, or correct it so you can move on. $\endgroup$
    – amWhy
    Dec 2 '18 at 19:00

We have $x_1= -x_2+x_3+x_4$.

Let $x_2=s$, $x_3=u$, $x_4=v$ (the free variables), then we have


$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=s\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + u\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + v\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$

  • $\begingroup$ Thank you for the answer. Also, for Ax=b, if I do row reduced echolen form for "b" size too, I get "2".. so same way we will do it for Ax=b? I am wondering what will be the particular solution and special solution? $\endgroup$
    – Ella
    Dec 2 '18 at 9:46
  • $\begingroup$ I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution. $\endgroup$ Dec 2 '18 at 9:52
  • $\begingroup$ No, for instance after row reduced echelon from I have "b" as 2 0 0 0 Then as x2,x3,x4 are free variables so we can set any value for them. This way we will have so many solutions? How to find all the solutions for Ax=b then? $\endgroup$
    – Ella
    Dec 2 '18 at 9:59
  • $\begingroup$ We find a particualr solution plus the general homogeneous solution part. $\endgroup$ Dec 2 '18 at 10:00
  • $\begingroup$ Could you help me in finding that? $\endgroup$
    – Ella
    Dec 2 '18 at 10:05

You have already found all solutions: it's the set of all $(x_1,x_2,x_3,x_4)$ such that $x_1+x_2-x_3-x_4=0$.


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