Given the following SLE:
\begin{cases} x + 2y + -2z +w =0 \\ x +2y - 3w=0\end{cases} Check if its solution set is a subspace find its basis.
Reason I posted here is that it's the first time I get such a question and I am not sure of my methodology.
What I did:
I set-up the following coefficients matrix: \begin{equation*} A= \begin{pmatrix} 1& 2 & -2 & 1 \\ 1 & 2 & 0& -3 \end{pmatrix} \end{equation*}
This can be reduce into:
\begin{equation*} A= \begin{pmatrix} 1& 2 & -2 & 1 \\ 0 & 0 & 1& -2 \end{pmatrix} \end{equation*}
So we get the following: $x+2y-2z+w=0$ and $y-2w=0 \Longrightarrow y=2w$, so in the first equation becomes $x-2z+5w=0$, any vectors $(x,y,z,w)$ with $x-2z+5w=0$ will still hold this property after vector addition and scalar multiplication hence they form a vector subspace of $R^4$.
Concerning the second part, a typical vector in this subspace is of the form $z(2,0,1,0)+w(-5,2,0,1)$ so these 2 vectors form a basis.
Is my way of approaching the problem correct, if no what else to do?
I think you have a little miscalculation, the rest of your reasoning is otherwise correct. After reducing the matrix your two equations read
\begin{align} x + 2y - 2z + w &= 0, \\ z- 2w &= 0. \end{align}
(Note the $z$ in the second equation.)
This implies $z = 2w$ and therefore $x + 2y -3w = 0$. You can now choose $y$ arbitrary and get $x = -2y + 3w$. Hence, every vector in the solution space can be represented as $(-2y + 3w, y, 2w, w)$. Now we have
$$(-2y + 3w, y, 2w, w) = y(-2, 1, 0, 0) + w(3, 0, 2, 1), $$
hence a basis is $\{ (-2, 1, 0, 0), (3, 0, 2, 1) \}$.
One little remark concerning notation: The row-reduced echelon form of $A$ should not be called $A$ again, since it is not the same matrix. They are only equivalent.