# Thread: Finding Conditions for a System To Be Inconsistent

1. ## Finding Conditions for a System To Be Inconsistent

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.

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

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

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!

2. ## Re: Finding Conditions for a System To Be Inconsistent

I don't know that this will help but I'd do it a different way. Your 2 equations determine two lines. For the system to have no solution the lines must be parallel but unequal. So their slope must be the same but their y intercepts different.

$\displaystyle y=\frac{ax - c}{b}$ and $y=\frac{d-bx}{a}$

the slopes are respectively $\displaystyle \frac{a}{b}$ and $\frac{-b}{a}$ so

$\displaystyle \frac{a}{b}=\frac{-b}{a}\Rightarrow a^2+b^2=0$

the intercepts are

$\displaystyle \frac{-c}{b}$ and $\frac{d}{a}$ so

$\displaystyle \frac{-c}{b} \neq \frac{d}{a} \Rightarrow (bd+ac)\neq 0$

3. ## Re: Finding Conditions for a System To Be Inconsistent

Thanks for the quick response! Your solution is certainly a whole lot more intuitive than mine!

Though, I'm still curious if forcing indeterminacy is a valid way to "break the system."

4. ## Re: Finding Conditions for a System To Be Inconsistent

Consider this as a matrix equation:

A(x,y) = (c,d).

The matrix A is clearly invertible UNLESS a = b = 0 (more on that later).

If that is so, then then A-1(c,d) provides a solution, so we are back to what I promised earlier: the case where a = b = 0.

This then becomes 0(x,y) = (c,d).

So the required conditions for there NOT to be a solution is:

(a = b = 0) AND (either c ≠ 0 or d ≠ 0 (or both)).

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