I am trying to solve this, but I am stuck....
1 2 -1
2 3 c
0 c -15
1 2 -1
0 -1 (2+c)
0 c -15
1 2 -1
0 1 (-2)-c))
0 1/15c 1
Hello, mathproblems!
Your method should work if you know what to look for.
(But from your responses, I'm not sure you do.)
$\displaystyle \text{Find }c\text{ so that A is not invertible: }\;A \;=\;\begin{vmatrix}1&2&\text{-}1 \\ 2&3&c \\ 0&c&\text{-}15 \end{vmatrix}$
We have: .$\displaystyle \left|\begin{array}{ccc}1&2&\text{-}1 \\ 2&3&c \\ 0&c&\text{-}15 \end{array}\right|$
$\displaystyle \begin{array}{c} \\ R_2-2R_1 \\ \\ \end{array}\left|\begin{array}{ccc}1&2&\text{-}1 \\ 0&\text{-}1&c\!+\!2 \\ 0&c&\text{-}15 \end{array}\right|$
$\displaystyle \begin{array}{c}R_1+2R_2 \\ \\ R_3+c\!\cdot\!R_2 \end{array}\left|\begin{array}{ccc} 1&0&2c+3 \\ 0&\text{-}1 & c+2 \\ 0&0& c^2\!+\!2c\!-\!15 \end{array}\right|$
A matrix is not invertible if an entire row is composed of zeros.
That is: .$\displaystyle c^2+2c-15 \:=\:0 \quad\Rightarrow\quad (x-3)(x+5) \:=\:0$
Therefore: .$\displaystyle c \,=\,3,\,\text{-}5$