$$ \left[ \begin{array}{ccc|c} 3m&m+12&m-3&3\\ 3&m-3&1&3m\\ 1&1&3&2 \end{array} \right] $$
For the above matrix, I would like to know how to solve for the value(s) of m for which the system has more than one solution.
I have turned it into reduced row echelon form, which is
$$ \left[ \begin{array}{ccc|c} 1&0&0&{-3(41m+46)\over (m-6)(8m+19)}-1\\ 0&1&0&{3(8m^2+3m-14)\over (m-6)(8m+19)}\\ 0&0&1&{9\over 8m+19} \end{array} \right] $$
Is there even value(s) of m for which the system has more than one solution?
If there isn't, would I write the answer as "no values of m"?
Thanks so much for your help
Guide:
Be careful when you divide a number, you could have divided by a zero.
You should examine what happens when $8m+19=0$ and also what happens when $m=6$.
Edit:
$$ \left[ \begin{array}{ccc|c} 3m&m+12&m-3&3\\ 3&m-3&1&3m\\ 1&1&3&2 \end{array} \right] $$
reduces to
$$ \left[ \begin{array}{ccc|c} 0&-2(m-6)&-8m-3&3-6m\\ 0&m-6&-8&3m-6\\ 1&1&3&2 \end{array} \right] $$
reduces to
$$ \left[ \begin{array}{ccc|c} 0&0&-8m-19&-9\\ 0&m-6&-8&3m-6\\ 1&1&3&2 \end{array} \right] $$
The only possible values for it to have more than one solution is when either $m=6$ or when $m=-\frac{19}{8}$.
Notice that when $m=-\frac{19}{8}$, it has no solution.
When $m=6$, the system becomes
$$ \left[ \begin{array}{ccc|c} 0&0&-67&-9\\ 0&0&-8&12\\ 1&1&3&2 \end{array} \right] $$
and we can see that it has no solution as well.