Show that the following system of equations has a solution for any value of the constant $\lambda$, using matrix method. \begin{cases} & x+2y+4z = 4 \\& 2x+3y+6z = 0 \\& \space{~~~~~~~}\lambda y+z = 4&\end{cases} Write all solutions in vector form. Is the solution unique for any given $\lambda$? Interpret your answer geometrically.
Its matrix form is $\begin{pmatrix} 1&2&4 &4 \\ 2&3&6 &0 \\ 0&\lambda &1 &4 \end{pmatrix}$ and this is row-equivalent to:
$ \begin{pmatrix} 1&2&4 &4 \\ 0&1&2 &8 \\ 0& 0 &1-2\lambda &4(1-2\lambda) \end{pmatrix} \to \begin{pmatrix} 1&2&4 &4 \\ 0&1&2 &8 \\ 0& 0 &1 &4 \end{pmatrix} \to \begin{pmatrix} 1&0&0 &-12 \\ 0&1&2 &8 \\ 0& 0 &0 &4 \end{pmatrix} \to \begin{pmatrix} 1&0&0 &-12 \\ 0&1&0 &0 \\ 0& 0 &1 &4 \end{pmatrix}$
I think if $\lambda = \frac{1}{2}$ we have infinite number of solutions and $(x,y,z) = (-12,0,4)$ otherwise. Hence it has a solution for any given $\lambda$. How do I interpret this geometrically?
Suppose $\;\lambda=\frac12\;$ , so you get the reduced matrix
$$\begin{pmatrix}1&2&4&4\\0&1&2&8\end{pmatrix}$$
from where you can give the general parametric solution in vectorial form:
$$y+2z\stackrel{2nd\text{ line}}=8\implies y=8-2z\;,\;\;x+2(8-2z)+4z=4\implies x=-12\implies$$
$$P:=\left\{\;\begin{pmatrix}-12\\8-2z\\z\end{pmatrix}\;:\;\;z\in\Bbb F\;\right\}=\begin{pmatrix}-12\\8\\0\end{pmatrix}+P_0$$
where $\;P_0\;$ is the solution space for the correspondent homogeneous system and the vector
$\;(-12,\,8,\,0)^t\;$ is a particular solution to the given non-homogeneous system.
You can also check that in fact
$$P_0=\left\{\;\begin{pmatrix}0\\-2z\\z\end{pmatrix}\;:\;\;z\in\Bbb F\right\}$$
and $\;\Bbb F\;$ is the field of definition of the system