Determining whether a subset is a subspace The problems are as follows:

Determine whether the following subsets of $\mathbb{R}^3$ are subspaces of $\mathbb{R}^3$.

*

*$A = \{(u^2, v^2, w^2) \,|\, u, v, w \in \mathbb{R} \}$,

*$B = \left\{(a, b, c) \,|\,
\begin{pmatrix}
a & b & c\\
1 & 2 & 0\\
0 & 1 & 2
\end{pmatrix} \text{is not invertible}\right\}$,

*$C = \left\{(x, y, z) \,|\, \begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\right\}$.


Here's what I've tried so far:


*

*$A$ is a subspace of $\mathbb{R}^3$ as it contains the $0$ vector (?).


*The matrix is not invertible, meaning that the determinant is equal to $0$. With this in mind, computing the determinant of the matrix yields $4a - 2b + c = 0$. The original subset can thus be represented as $B = \left\{\left(\frac{2s - t}{4}, s, t\right) \,|\, s, t \in \mathbb{R}\right\}$; i.e. $B = \text{span}\left\{(\frac{1}{2}, 1, 0), (-\frac{1}{4}, 0, 1)\right\}$, a plane in $\mathbb{R}^3$.


*Solving for the linear system,
$$
\begin{aligned}
\begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} &=
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
5\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
9
\end{pmatrix}
&= 
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
-1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
3\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
7
\end{pmatrix}
&= 
\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}
\end{aligned}$$
Converting to row echelon form gives the trivial solution $x = 0, y = 0, \text{ and } z = 0$; $C$ only contains the $0$ vector.

Is my reasoning correct? Also, quite a bit of the computations have been omitted in this post for brevity. Regardless of this, are my answers well justified?
Thank you.

You proof seems fine to me but for case $A$ we should observe that in general
$$(u_1^2, v_1^2, w_1^2) + (u_2^2, v_2^2, w_2^2)\neq ((u_1+u_2)^2, (v_1+v_2)^2, (w_1+w_2)^2)$$
and you can easily find by yourself an example where equality doesn't hold which is a necessary step to complete the proof.
