[Math] When does a system of equations have no solution

linear algebramatrices

I have performed Gaussian elimination on this matrix to reduce it to

$$
\left[
\begin{array}{@{}ccc|c@{}}
-3&-1&2 & 1 \\
0& \frac{-5}{3}& \frac{10}{3} & \frac{8}{3} \\
0&0& a+2 & b + \frac{6}{5} \\
\end{array}
\right]
$$

I thought that setting $a$ equal to $-2$ and having $b$ not equal to $-\frac{6}{5}$ would be the answer to this problem, but it apparently isn't. Could someone please explain why?

Best Answer

there is no solution when the matrix is $\textbf{inconsistent}$. This means you will have a zero row in your reduced matrix corresponding to a non-zero entry of the desired solution eg.

$$ \left[ \begin{array}{@{}ccc|c@{}} -3&-1&2 & 1 \\ 0& \frac{-5}{3}& \frac{10}{3} & \frac{8}{3} \\ 0&0& 0 & \text{any non-zero} \\ \end{array} \right] $$

this is because the third row would imply $0*x+0*y+0*z = 0 = c \ne 0$ which is obviously false