Use determinants to find which real values of c make each of the following matrices invertible So i was given this question 
Use determinants to find which real values of c make each of the following matrices invertible
$
\left[ {\begin{array}{cc}
0 & c & -c \\
-1 & 2& 1 \\
c & -c & -c
\end{array} } \right]
$
When i look at the solutions for this question they usually adding or subtracting a column or row to one another, until there is two consecutive $0s$ in the row or column then make a 2x2 matrix. I understand the logic behind finding a determinant, but for this question and similar ones is there a rule or method to go about solving these kinds of problems, and how to do it?
 A: If a given square matrix is invertible, then determinant of it should be non zero. The determinant of this matrix is $-c(c-c)-c(c-2c)=c^2$. So, any non zero $c$ would make this matrix invertible.
A: Note that if you multiply a row or column by a constant, then since the
determinant is multi linear, the determinant of the resulting matrix is a constant times the original matrix.
Hence 
$\det \begin{bmatrix}
0 & c & -c \\
-1 & 2& 1 \\
c & -c & -c
\end{bmatrix} = c 
\det \begin{bmatrix}
0 & 1 & -1 \\
-1 & 2& 1 \\
c & -c & -c
\end{bmatrix} = c^2 
\det \begin{bmatrix}
0 & 1 & -1 \\
-1 & 2& 1 \\
1 & -1 & -1
\end{bmatrix}$.
We also have $\det \begin{bmatrix}
0 & 1 & -1 \\
-1 & 2& 1 \\
1 & -1 & -1
\end{bmatrix} = 1$.
A: First, by multilinearity, you have:
$$\begin{vmatrix}
0 & c & -c \\
-1 & 2& 1 \\
c & -c & -c
\end{vmatrix}
=c^2\begin{vmatrix}
0 & 1 & -1 \\
-1 & 2& 1 \\
1 & -1 & -1
\end{vmatrix}
$$
Then, with row or column operations:
$$\begin{vmatrix}
0 & 1 & -1 \\
-1 & 2& 1 \\
1 & -1 & -1
\end{vmatrix}=\begin{vmatrix}
0 & 1 & -1 \\
0 & 1& 0 \\
1 & -1 & -1
\end{vmatrix}=1$$
(develop along the first column). Thus the determinant is equal to $c^2$, and it is non-zero if and only if $c\neq0$.
