How to prove infinite solution vs no solution for singular matrix problem. In the problem Ax=B
My coefficient matrix is
\begin{bmatrix}
        α-1 & 1-α\\
        α & -α
        \end{bmatrix}
x is
\begin{bmatrix}
        ln K\\
        ln L
        \end{bmatrix}
b is
\begin{bmatrix}
        ln (r/α) - ln (pA)\\
        ln (w/(1-α)) - ln(pA)
        \end{bmatrix}
I'm aware the matrix is singular and therefore there is no unique solution, however I'm informed from the solution of the problem set that there are infinitely many solutions if $$α ln(r/α) + (1 − α) ln(w/(1 − α)) = ln(pA)$$ and no solution otherwise.
I am unclear how to so come to this conclusion. Any help appreciated.
Cheer
 A: Before I start, I assume that the first $A$ is a matrix and the second $A$ (in the logarithm) is a parameter, I will denote it $a$ so $\log(p A)$ becomes $\log (p a)$.
To say there exists no solution means that the two equations (read as the rows of the equation $Ax=b$) are inconsistent. It is easy to find (from one of the two equations) the answer $\log (K)= \frac{1}{\alpha}\left(\alpha \log(L)-\log(ap) + \log \left(\frac{w}{1-\alpha}\right)\right)$. Substituting this into the second equation we find that $\log(L)$ cancels (precisely because det$A = 0$), so we cannot solve for $\log(L)$, however the equation which is left is precisely the one which you gave in your question. This means that if that equation does not hold then the equations are inconsistent and there does not exist a solution, otherwise there exists (infinitely many) solution(s). 
I would recommend studying at Linear Algebra. Consider the matrix equation $A \left(\begin{array}{c} x \\ x \end{array} \right) = b$ (for some $A$ and $b$), this defines two straight lines on the plane $\mathbb{R}^{2}$. The statement that there is no solution for $x,y$ to this matrix equation can be geometrically interpetted as saying that the straight lines do not cross each other. I think this is useful in understanding what is happening. 
