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As an extension to Finding null space of matrix. which asked about the nullspace of the rref matrix: $$R = \begin{bmatrix} 1 & -2 & 0 & -1 & 3 \\ 0 & 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

I wanted to generate a particular solution as well, using some $d = \begin{bmatrix} 3\\ 1\\ 0\\ \end{bmatrix}$ with $Rx = d$. The $d$ I chose was random, with the exception that the last row had to contain $0$ (or $d$ wouldn't be in the column space of $R$). I solved this system by setting all the free variables ($x_2$, $x_4$, $x_5$) to $0$ and then solving for the pivot variables ($x_1$, $x_3$), which gave me $x_1 = 3, x_3 = 1$, and thus $$x_p = \begin{bmatrix} 3\\ 0\\ 1\\ 0\\ 0\\ \end{bmatrix}$$ So the final solution seems to be the superposition of the particular part and the homogeneous part, which is:

$$x=\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ \end{bmatrix} = \begin{bmatrix} 3\\ 0\\ 1\\ 0\\ 0\\ \end{bmatrix} + x_2 \begin{bmatrix} 2\\ 1\\ 0\\ 0\\ 0\\ \end{bmatrix} + x_4 \begin{bmatrix} 1\\ 0\\ -2\\ 1\\ 0\\ \end{bmatrix} + x_5 \begin{bmatrix} -3\\ 0\\ 2\\ 0\\ 1\\ \end{bmatrix}$$

Is this the correct thought process when coming up with a solution to a nonhomogeneous problem? Also, why is it necessary to set the free variables to 0 when solving for the pivot variables? Thanks!

Raleigh Clemens
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  • The solution set is always given as $x_p + N(A)$. 2. Setting the free variables to $0$ gives you a particular solution. 3. Here $\text{rank($A$)} = 2$ and the column space $C(A)$ is contained in the $xy$-plane. Thus in fact $C(A)$ is the $xy$-plane, and any vector $d$ with third component zero will be solvable.

    • Note: I have been very terse. Let me know if any facts need clarification.

    – twosigma Jan 10 '21 at 14:31
  • Thank you for the answer! I was wondering why fact (2) is true- why does setting the free variables to 0 allow you to find the particular solution? What does this mean intuitively (or physically) about the solution to the system if the free variables are set to 0? – Raleigh Clemens Jan 10 '21 at 17:28
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    $Ax$ is a linear combination of the columns of $A$, with the coefficients coming from $x$. Therefore if $A$ is in RREF, we can set all free variables to $0$ and set each leading variable to the corresponding entry in $d$ (like you did in your example), and we will get a particular solution to $Ax = d$. I am not sure about the intuition, but this is a convenient way of achieving a particular solution. If we know $N(A)$, then finding any particular solution (as we have just done) completely determines the solution set of $Ax = d$. – twosigma Jan 10 '21 at 18:59