<s>[INST] 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?
 [/INST] 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.

</s>