Hello everyone. I've been trying to come up with an answer to this question for a while now, however, I haven't had much luck.

My solution (first by reducing the matrix with elementary matrix operations):Find conditions on a, b, c, and d such that $\displaystyle ax-by=c$ and $\displaystyle bx+ay=d$ has no solution.

$\displaystyle \begin{bmatrix}a & -b & c \\b & a & d\end{bmatrix}\Rightarrow\begin{bmatrix}a & 0 & c+\frac{b(ad-bc)}{a^2+b^2} \\0 & b & \frac{b(ad-bc)}{a^2+b^2}\end{bmatrix}$

I know that a system will be inconsistent if a row (in this case) takes the form of this vector:

$\displaystyle \vec{R}_n = \begin{bmatrix}0, & 0, & a\end{bmatrix}, a \in \mathbb{R}_{\neq 0}$

Therefore, for the system to be inconsistent, it looks like $\displaystyle a$, $\displaystyle b$, or both should have to be zero. However, if $\displaystyle a=0$, the first row becomes the zero vector. If $\displaystyle b=0$, I get the zero vector in the second row. If I set both $\displaystyle a$ and $\displaystyle b$ equal to zero, I end up with an answer that doesn't quite make sense to me (because of the indeterminate 0/0 terms):

$\displaystyle \begin{bmatrix}0 & 0 & c+?\\ 0 & 0 & ?\end{bmatrix}$

Now, if $\displaystyle c \in \mathbb{R}_{\neq 0}$, the condition will be met, but I don't know how to reconcile the indeterminacy generated by the 0/0 quotients. Help would be greatly appreciated!