<s>[INST] Check if vectors are the the fundamental set of solutions $$\begin{cases} x_1+2x_2+x_3-x_4+x_5=0 \\ 2x_1+x_2-x_3+2x_4-x_5=0 \\ x_1+5x_2+4x_3-5x_4+4x_5=0 \\ 4x_1+5x_2+x_3+x_5=0 \end{cases}$$
$$\begin{cases} \vec{a}_1=(-5,4,3,3,-3) \\ \vec{a}_2=(-7,5,-3,6,6) \\ \vec{a}_3=(5,-4,6,-3,-6) \end{cases}$$
Solution:
\begin{bmatrix}
    1&2&1&-1&1 \\
    2&1&-1&2&-1 \\
    1&5&4&-5&4 \\
    4&5&1&0&1
\end{bmatrix}
Reducing to the row echelon form gives us
\begin{bmatrix}
    1&2&1&-1&1 \\
    0&3&-3&-4&-3
\end{bmatrix}
Let $x_3 = c_1$, $x_4=c_2$, $x_5=c_3$.
$$\begin{cases} x_1+2x_2+c_1-c_2+c_3=0 \\ 0x_1+3x_2-3c_1-4c_2-3c_3=0\end{cases}$$
$$x_2=c_1+\frac{4}{3}c_2+c_3$$
$$x_1=-2c_1-\frac{8}{3}c_2-2c_3-2c_1+c_2-c_3=-4c_1-\frac{5}{3}c_2-3c_2$$
However, plugging the corresponding values of each vector into the resulting system does not work here. What did I miss?
 [/INST] $$
\left(
\begin{array}{rrrr}
 1 & 0 &  0 &0 \\
-1 & 0 & 1 &0 \\
 3 &-1 &-1 &0 \\
-5 & 0 & 1& 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1& 2  &1& -1 & 1\\
2& 1 &-1 & 2 &-1\\
1 &5 & 4 &-5 & 4\\
4 &5&  1 & 0&  1\\
\end{array}
\right) =
\left(
\begin{array}{rrrrr}
1& 2  &1& -1 & 1\\
0& 3 &3 & -4 &3\\
0 &0 & 0 &0 & 0\\
0 &0&  0 & 0&  0\\
\end{array}
\right)
$$
</s>