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I'm stuck at part b) of the following problem:

a) Find a $2\times 3$ system $Ax = b$ whose complete solution is $$\vec x = \begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix} + w \begin{bmatrix} 1\\ 3\\ 1 \end{bmatrix}.$$

b) Find a $3\times 3$ system with these solutions exactly when $b_1 +b_2 = b_3$.

(Linear Algebra and Its Applications by Gilbert Strang - Chapter 2.2 problem 10)

For part a) I arrived at the following solution: $$\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -3 \end{bmatrix} \vec x = \begin{bmatrix} 1\\ 2 \end{bmatrix} \quad \quad (R \vec x = \vec b)$$ by "reading of" the entries for $x_1$ and $x_2$ from the nullspace $w\begin{bmatrix} 1\\ 3\\ 1 \end{bmatrix}$ and putting them into a row reduced echelon form matrix $R$ with switched signs and by again reading of the values for $b_1$ and $b_2$ from the particular solution $\begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix}$.

For part b) I thought that I could do the same:

The matrix $R=\begin{bmatrix} 1 & 0 &-1\\ 0 & 1 &-3\\ 0 & 0 &0 \end{bmatrix}$ has $N(R) = w\begin{bmatrix} 1\\ 3\\ 0 \end{bmatrix}$. Now I wanted to solve for $\vec b$:

$$\begin{bmatrix} 1 & 0 &-1\\ 0 & 1 &-3\\ 0 & 0 &0 \end{bmatrix} \vec x =\begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix} $$ $$\begin{bmatrix} 1 & 0 &-1\\ 0 & 1 &-3\\ 0 & 0 &0 \end{bmatrix} \vec x =\begin{bmatrix} b_1\\ b_2\\ b_1+b_2 \end{bmatrix} $$ Since $\vec x = \begin{bmatrix} 1\\ 2\\ 0 \end{bmatrix}$ is a particular solution: $b_1 = 1$,$b_2 = 2$. However: $b_1 + b_2 = 3$ which makes the system inconsistent. Is there a system with the given solution which is consistent? And if there is one, where did I make a mistake?

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Suppose the matrix $A=\begin{bmatrix}\mathbf{3u} & \mathbf{-(u+v)} & 3\mathbf{v}\end{bmatrix}$, where $\bf{u}$ and $\bf{v}$ are $3 \times 1$ column vectors. The reason for starting with this $A$ is to make sure that the null space of $A$ is spanned by $(1,3,1)$ (as required by the problem). (Note: of course we want to make sure that $\bf{u}$ and $\bf{v}$ are independent so that null space is $1-$dimensional.)

Now we want $(1,2,0)$ to be a solution for $Ax=\begin{bmatrix}b_1\\b_2\\b_1+b_2\end{bmatrix}$. This means we want $$1(\mathbf{3u})-2 \mathbf{(u+v)}+0\mathbf{(3v)}=\begin{bmatrix}b_1\\b_2\\b_1+b_2\end{bmatrix}.$$ This is equivalent to $$\mathbf{u}-2 \mathbf{v}=\begin{bmatrix}b_1\\b_2\\b_1+b_2\end{bmatrix}.$$ So for example, if we have $$\mathbf{u}=\begin{bmatrix}b_1\\b_2\\0\end{bmatrix},$$ then we can have $$\mathbf{v}=\begin{bmatrix}0\\0\\-\frac{b_1+b_2}{2}\end{bmatrix}.$$ Now you have your $A$. If you assume that at least one of $b_1$ or $b_2$ is non-zero, and $b_1+b_2 \neq 0$, then the conditions are (easily) satisfied.

Your error: is that while you are row reducing the matrix $A$ but you are not applying the row operations to the right hand side vector $\bf{b}$.

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  • $\begingroup$ Okay, I understand your solution but I am still unsure about where I made a mistake because I do not really row reduce anything on the left but simply start with a reduced matrix, do I? $\endgroup$
    – David A.
    Jun 8, 2021 at 20:35
  • $\begingroup$ @DavidA. When you start with a row reduced matrix $R$ and then you write $Rx=b$, for consistency you will have to assume that $b_3=0$, said differently you will have to assume that $b_1+b_2=0$. This means you cannot choose arbitrary $b_1,b_2$ for the system. That is why I said that when you work with $R$ instead of $A$, you have performed the row operations on $A$ but not not on $\bf{b}$. $\endgroup$
    – Anurag A
    Jun 8, 2021 at 20:40
  • $\begingroup$ @DavidA. To add further to my comment, say you want to solve for $Ax=b$. But if you start with a row reduced matrix $R$ such that the null space of $R$ (hence that of $A$) is one dimensional, then the last row of $R$ has to be a zero row (hence exactly two pivots). This means for any right side $b$ that has non-zero last entry cannot be a "good" right side for $Rx=b$, even though it may very well be a "good" right side for $Ax=b$. Hope this clears the dilemma. $\endgroup$
    – Anurag A
    Jun 8, 2021 at 21:05
  • $\begingroup$ @DavidA. So if you want to construct an example using your idea, then make sure that you choose $b_1$and $b_2$ such that $b_1+b_2=0$ and your example will work without any issues. $\endgroup$
    – Anurag A
    Jun 8, 2021 at 21:08
  • $\begingroup$ All right, the clarification and elaboration cleared up the confusion I had. Thank you for your help :D $\endgroup$
    – David A.
    Jun 9, 2021 at 15:48

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