Different solutions to the same system of linear equations I want to solve a real homogeneous system of linear equations represented by this matrix:
\begin{bmatrix}0&2&2&7&1&0\\1&0&1&3&1&0\\-1&2&1&4&0&0\end{bmatrix}
The reduced echelon form is:
\begin{bmatrix}1&0&1&3&1&0\\0&1&1&7/2&1/2&0\\0&0&0&0&0&0\end{bmatrix}
Since I have used Maple to verify my answer, I know that the above matrix is correct. Now comes my question. When deriving the parameterized solution from the reduced echelon form, I want to define the free variables as:
\begin{align*}
x3=t1
\end{align*}
\begin{align*}
x4=t2
\end{align*}
\begin{align*}
x5=t3
\end{align*}
Then the solution has to satisfy the following equations:
\begin{align*}
x1+t1+3t2+t3=0 \leftrightarrow\ x1=-t1-3t2-t3
\end{align*}
\begin{align*}
x2+t1+(7/2)t2+(1/2)t3=0 \leftrightarrow\ x2=-t1-(7/2)t2-(1/2)t3
\end{align*}
Thus the parameterized solution satisfy:
\begin{align*}
\begin{bmatrix}x1\\x2\\x3\\x4\\x5\end{bmatrix}=t1\begin{bmatrix}-1\\-1\\1\\0\\0\end{bmatrix}+t2\begin{bmatrix}-3\\-7/2\\0\\1\\0\end{bmatrix}+t3\begin{bmatrix}-1\\-1/2\\0\\0\\1\end{bmatrix}t1,t2,t3\in\mathbb{R}
\end{align*}
However, when I solve the system in Maple, I get this:
\begin{align*}
\begin{bmatrix}x1\\x2\\x3\\x4\\x5\end{bmatrix}=t1\begin{bmatrix}2\\1\\0\\0\\-2\end{bmatrix}+t2\begin{bmatrix}1\\0\\1\\0\\-2\end{bmatrix}+t3\begin{bmatrix}4\\0\\0\\1\\-7\end{bmatrix}t1,t2,t3\in\mathbb{R}
\end{align*}
Clearly, this is the solution I will get if I define the free variables as:
\begin{align*}
x2=t1
\end{align*}
\begin{align*}
x3=t2
\end{align*}
\begin{align*}
x4=t3
\end{align*}
Since the solution then has to satisfy:
\begin{align*}
t1+t2+(7/2)t3+(1/2)x5=0 \leftrightarrow\ x5=-2t1-2t2-7t3
\end{align*}
\begin{align*}
x1+t2+3t3+x5=0 \leftrightarrow\ x1=-t2-3t3-x5=-t2-3t3-(-2t1-2t2-7t3)=2t1+t2+4t3
\end{align*}
Are both solutions correct? If yes, does it mean that the three vectors in the two different solutions respectively spans the same vector space with respect to different bases? Also, I would really like to know why Maple does this? What is the advantage of choosing other free variables than those you normally choose?

It is easy to make Maple to use your choices of free variables, how? Let us rewrite what you are looking for mathematically. What you are asking can be written as a new linear system.
$$\begin{bmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
1 & 0 & 1 & 3 & 1\\
0 & 1 & 1 & \frac{7}{2} & \frac{1}{2}\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{bmatrix}=\begin{bmatrix}
t_1\\
t_2\\
t_3\\
0\\
0\\
0
\end{bmatrix}$$
Then the Maple code using the same LinearSolve command of the LinearAlgebra package that you have used is the following.
A := Matrix([ [0,0,1,0,0], [0,0,0,1,0], [0,0,0,0,1], [1,0,1,3,1], [0,1,1,7/2,1/2], [0,0,0,0,0] ]);
b := < t__1, t__2, t__3, 0, 0, 0 >;
LinearAlgebra:-LinearSolve( A, b );

The output is exactly what you found manually.

