Quadratic equation not equal to zero (solving a matrix with a parameter) I came across this in my matrix module, learning about number of solutions when the matrix has parameters.
$$
\left[
\begin{array}{ccc|c}
1 & -1&1&1\\0&2&k^2&k^2-4\\0&0&k^2+k-2&k^2-4
\end{array}
\right]
$$
For there to be one solution, I need $k^2+k-2\ne0$, so that number of rows equal the number of unknowns.  I came up with $(k+2)(k-1)\ne0$ therefore solutions $k\ne-2$ and $k\ne1$.  But the worked solution to the question I'm doing has solutions $k\ne2$ and $k\ne-1$ (same values, but sign reversed).
Is there something I've missed when solving an inequation, or is there a mistake in the solution?
 A: Factor $k^2-4$ as $(k+2)(k-2)$. You can see that, in conjunction with $(k+2)(k-1)$ that when $k=-2$, both expressions are zero, this results in a consistent system; the other option (as a result of factoring  $k^2-4$, is for $k=2$. 
And if I am not mistaken, you are correct about the second solution, $k=1$. I think there was a solution error for that.
Also, you are not finding ONE solution. This system, a 3X4, will have infinitely many, or none. What it is asking for are values of $k$ which make the system inconsistent. And, try to avoid doing math with the not equals symbol. It is generally a good way to write a solution, but it is often much more confusing and can lead you down the wrong path. 
A: The fourth column is an augmented matrix, and no, that does not change my answer. Namely, because $k$ is included in the augmented column. If a value of $k$ makes that entire bottom row zero, including the augmented column entry, then there can be solutions between the first two rows. If such a value of $k$ made the bottom row zero, then it is consistent: $a*0+b*0+c*0=0$ for all $a, b, c$. 
