Different solutions to the same system of linear equations

Question

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?

Answer 1

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.