Linear system resolution

Question

I've got this linear system \begin{cases} x-3y+6z=0 \\ kx-3y+6z=0\end{cases} I must to find the basis and the size of this linear system. I have a stuck becose here are $kx$. I made it so: $$\begin{align} x&=3y-6z\\ k(3y-6z)&=3y-6z\end{align}$$But I see that my method it's wrong in this case. Cos' I don't know how to solve it. Is someone who can tell to me whot can I do in this case? Or whot I must to study? I try to search on Google but I don't find anything similar.

Answer 1

The second equation may be written as $(k-1)(3y-6z)=0$. Now if $k=1$ you have $x=3y-6z$, hence $(3,1,0)$ and $(-6,0,1)$ form a basis. otherwise $3y-6z=0$, which implies $x=0$. Here $(0,2,1)$ is a basis vector.

Answer 2

The matrix is \begin{bmatrix} 1 & -3 & 6 \\ k & -3 & 6 \end{bmatrix} that easily becomes \begin{bmatrix} 1 & -3 & 6 \\ k-1 & 0 & 0 \end{bmatrix} Now there are two cases: if $k=1$, only one equation remains, namely $x=3y-6z$, so a basis is $$ \left\{\, \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -6 \\ 1 \\ 1 \end{bmatrix} \,\right\} $$ If $k\ne1$, there are two equations, $x=0$ and $y=2z$; the RREF is \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \end{bmatrix} Therefore the basis is $$ \left\{\, \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} \,\right\} $$

Answer 3

Ok thanks I made it. $$\begin{bmatrix} 1 & - 3 & 6 \\ k & - 3 & 6 \end {bmatrix}$$ Determinate "k". 1*(-3)-k*(-3)=-3+3k. k=1. For k=1 {(x=3y-6z):x,y €R} (3y,y,0),(-6z, 0,z) S=(3,1,0),(-6,0,1) generated W. So this is the basis and dimension =2. For k=0 -3y=-6z {(y=2z):z €R} (0,2z,z) S=(0,2,1) generated W, so this is a bases and the dimension =1.