# Linear algebra: $Ax=0$ for matrix

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 $$Ax=0$$?

• Hyperplan equation ! Dec 2 '18 at 9:42
• 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. 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

$$x_1=-s+u+v.$$

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

• 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?
– Ella
Dec 2 '18 at 9:46
• I am guessing you mean $x_1=2-x_2+x_3+x_4$. then $(2,0,0,0)$ is a particular solution. Dec 2 '18 at 9:52
• 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?
– Ella
Dec 2 '18 at 9:59
• We find a particualr solution plus the general homogeneous solution part. Dec 2 '18 at 10:00
• Could you help me in finding that?
– 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 $$x_1+x_2-x_3-x_4=0$$.