<s>[INST] Determining a basis for solution set 
For a given equation system (over field $R$), determine the basis for
  the space $\operatorname{Sol}(U)$ (where $\operatorname{Sol}$ is a set of solutions):
\begin{cases} 2x + 7y + 3z+t=0 \\ 3x + 5y + 2z + 2t=0 \\ 9x + 4y + z+7t=0 \end{cases}

It's a new type of problem for me and I'm struggling a bit. I know what a basis is but am quite confused about the $\operatorname{Sol}(U)$ part. Does that mean I should just try to solve the system and then think of such a basis that it spans over the possible solutions?
 [/INST] Put the system into an augmented matrix:
$$\left[\begin{array}{rrrr|r}
    2 & 7 & 3 & 1 & 0 \\
    3 & 5 & 2 & 2 & 0 \\
    9 & 4 & 1 & 7 & 0
  \end{array}\right]$$
Use Gauss-Jordan elimination to get it into reduced row echelon form.
$$\left[\begin{array}{rrrr|r}
    1 & 0 & -\frac{1}{11} & \frac{9}{11} & 0 \\
    0 & 1 & \frac{5}{11} & -\frac{1}{11} & 0 \\
    0 & 0 & 0 & 0 & 0
  \end{array}\right]$$
Now you have a set of equations:
$$\begin{align}
x+0y+-\frac{1}{11}z+\frac{9}{11}t&=0 \\
0x+y+\frac{5}{11}z-\frac{1}{11}t &= 0
\end{align}$$
Finally, parameterize it with $u$ and $v$ as follows:
$$\begin{align}
x&=\space\space\space\frac{1}{11}u-\frac{9}{11}v \\
y&=-\frac{5}{11}u+\frac{1}{11}v \\
z&=u \\
t&=v \end{align}$$
The span of the solution set is therefore:
$$
u\begin{bmatrix}
\frac{1}{11} \\
-\frac{5}{11} \\
1 \\
0
\end{bmatrix}+
v\begin{bmatrix}
-\frac{9}{11} \\
\frac{1}{11} \\
0 \\
1
\end{bmatrix}
$$
</s>