I'm given the matrix: \begin{bmatrix}-a-7&-2a-6 & 0\\a+3&a+3 &3a-3\\ -2a-14 & -4a-12 & a-1\end{bmatrix} and I wanted to determine which values of $a$ give the matrix only $1$ free variables, so I got the matrix into this echelon form: \begin{bmatrix}-a-7&-2a-6&0\\0&\left(\frac{(a+3)(1-a)}{a+7}\right)&3a-3 \\ 0 & 0 & a-1\end{bmatrix}
so clearly $a=1$ gives the matrix two free variables, so $3$ is not a solution,
and $a=-3$ gives the matrix a single free variable as desired, so $1$ is a solution.
Where I'm unclear is with $a=-7$ since that would make the top left entry $=0$, but it would also make the middle entry (second diagonal entry) be $(\frac{-32}{0})$ which is undefined since you cannot divide by zero. But when I let $a=-7$ in the original matrix, I get $0$ free variables.
Am I missing something?
Before we divide by a number, we check if it is zero and we can handle them separately. When $a=-7$.
The original matrix becomes
$$\begin{bmatrix} 0 & 8 & 0 \\ -4 & -4 & -24 \\ 0 & 16 & -8\end{bmatrix}$$ which is clearly a ful rank matrix, hence your computation for $a=-7$ is right.
Note that when you mean $1$ free variable, if you mean the number of free parameter in a linear system which is the nullity. Then when $a=1$, the rank of your matrix is $1$ and hence the nullity is $2$.