# System of equations from given solution

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?

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 $$\bf{b}$$.
• @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 $\bf{b}$. Jun 8, 2021 at 20:40
• @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. Jun 8, 2021 at 21:05
• @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. Jun 8, 2021 at 21:08