# Find $A$ if solution to $Ax=b$ is given

Solution to $$Ax=b$$ is $$x=\begin{bmatrix} 1\\ 0\\ 1\\ 0 \end{bmatrix}+\alpha_{1}\begin{bmatrix} 1\\ 1\\ -1\\ 0 \end{bmatrix}+\alpha_{2}\begin{bmatrix} 1\\ 0\\ 1\\ 1\end{bmatrix}$$. For $$b=(1,2,1)^T$$, find $$A$$.

Somehow, Gilbert Strang says that it is obvious that first and third column should add up to $$(1,2,1)^T$$, and then he says that second column is third minus first and fourth is -(first + third). I don't see how he concluded all of this, so any clarification is very welcome.

Note that the $$\alpha_i$$ given in the solution for $$x$$ are arbitrary. This means, that regardless of what values of $$\alpha_i$$ you have, the resulting $$x$$ is a solution for $$Ax = b$$.

Now, if we take $$\alpha_1 = \alpha_2 = 0$$, then we get that $$x = (1,0,1,0)$$ is a solution to $$Ax = b$$. If we look at how we multiply a matrix with a vector , then $$(Ax)_i = \sum_{j} A_{ij}x_j$$. Therefore, since $$x_2=x_4 = 0$$ and $$x_1=x_3 = 1$$, we get that $$(Ax)_i = A_{i1} + A_{i3}$$, for all $$i$$.

But this is equal to $$b_i$$, since $$Ax = b$$. Therefore, $$b_i = A_{i1} + A_{i3}$$, for all $$i$$.

Now, if you look at the first column vector $$A_{j1}$$ and the third column vector $$A_{j3}$$, the fact that $$b_i = A_{i1} + A_{i3}$$ for all $$i$$, is a restatement of the fact that $$b$$ is the sum of the first and third columns of $$A$$.

Similarly, use the fact that if $$x_1$$ and $$x_2$$ satisfy $$Ax_1 = Ax_2 = b$$ then $$A(x_1 - x_2) = 0$$ (zero vector). Now, see if you can take two different values of $$\alpha_{i}$$ for $$x_1$$ and $$x_2$$. Get a vector $$x_1 - x_2 = y$$ such that $$Ay = 0$$. Now, use the entries of $$y$$ , as I did above, to conclude certain relations about the columns of $$A$$. See what $$y$$ will give you the relations that Gilbert and Strang refer to.

You have that $$A \begin{bmatrix} 1+\alpha_1+\alpha_2 \\ \alpha_1 \\ 1-\alpha_1+\alpha_2 \\ \alpha_2 \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$$ If $$A=[a_1\ a_2\ a_3\ a_4]$$, this means that $$(1+\alpha_1+\alpha_2)a_1 +\alpha_1a_2 +(1-\alpha_1+\alpha_2)a_3 +\alpha_2 a_4 =\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=b$$ for every $$\alpha_1$$ and $$\alpha_2$$.

In particular you have $$\begin{cases} a_1+a_3=b & (\alpha_1=0, \alpha_2=0) \\[4px] 2a_1+a_2+a_3=b & (\alpha_1=1, \alpha_2=0) \\[4px] 2a_1+a_3+a_4=b & (\alpha_1=0, \alpha_2=1) \end{cases}$$ This is quite similar to a linear system, isn't it? $$\begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 2 & 1 & 1 & 0 & b \\ 2 & 0 & 1 & 1 & b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 0 & 1 & -1 & 0 & -b \\ 0 & 0 & -1 & 1 & -b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 0 & 1 & -1 & 0 & -b \\ 0 & 0 & -1 & 1 & -b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 & 0 & b \\ 0 & 1 & -1 & 0 & -b \\ 0 & 0 & 1 & -1 & b \\ \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & -1 & b \\ \end{bmatrix}$$ Thus $$a_1=-a_4$$, $$a_2=a_4$$, $$a_3=a_4+b$$. The fourth column can be anything; so the matrix $$A$$ is $$\begin{bmatrix} -x & x & x+b & x \end{bmatrix}$$ where $$x$$ is an arbitrary $$3\times 1$$ column.