How to understand this question and answer from MIT 18.06 linear algebra? This is one of the problems from the MIT 18.06 Linear Algebra course:


Problem 16.1:$\quad$ ($4.1\ \#7$. Introduction to Linear Algebra: Strang) For every system of $m$ equations with no solution, there are numbers $y_1,...,y_m$ that multiply the equations so they add up to $0 = 1$. This is called Fredholm's Alternative:
Exactly one of these problems has a solution: $A\mathbf x= b$ OR $A^T\mathbf y = 0$ with $\mathbf y^T\mathbf b = 1$.
If $\mathbf b$ is not in the column space of $A$ it is not orthogonal to the nullspace of $A^T$. Multiply the equations $x_1 – x_2 = 1, x_2 – x_3 = 1$ and $x_1 – x_3 = 1$ by numbers $y_1, y_2$ and $y_3$ chosen so that the equations add up to $0 = 1$.


I'm unsure how to go about even starting this question. Even the answer doesn't really make sense to me.


 A: It is a fact that $\operatorname{Null}(A^T)\perp\operatorname{Col}(A)$. So to say that $Ax=b$ has no solution, which is to say that $b$ is not in the column space of $A$, is to say that $b$ is not perpendicular to the nullspace of $A^T$. That is, there is a $y$ s.t. $A^Ty=0$ and so that $\langle b,y\rangle = y^Tb\neq 0$. Adjusting $y$ you can make this last non-zero number $1$. Note that $A^Ty=0$ is equivalent to $y^TA=0$, so if $Ax=b$ has no solution, then we can find $y$ such that $0=y^TAx$ and $y^Tb=1$. That is, we can find a linear combination of the equations of the system $Ax=b$ which read $0=1$. The problem ask us to produce the coefficients of this linear combination. I.e. to verify this principle on the particular unsolvable equation $Ax=b$ of the problem.
A: Let us understand this problem with a 3-dimensional space perspective.
$$y_1(x_1-x_2)  + y_2(x_2-x_3) + y_3(x_1-x_3) = y_1 + y_2 + y_3$$
On rearranging,
$$x_1(y_1 + y_3)  + x_2(y_2 - y_1) + x_3(-y_2-y_3) =  y_1 + y_2 + y_3$$
Since $y_1$, $y_2$ and $y_3$ are constants, the above is an equation of plane. See Wikipedia-Plane(geometry).
According to question, we need to end up with $0 = 1$, which is "not real", implies that the above equation should not be valid. That is, we cannot make a plane out of the equation. We need to find the values of the constants for which no plane can be formed out of the equation. It is clearly visible that it occurs when,
$$y_1=1,  y_2=1,  y_3=-1$$
