Not every permutation matrix has determinant $-1$, but the elementary matrices which are permutation matrices (corresponding to interchanges of two rows) have determinant $-1$. The easy way to see this is that (1) the identity matrix has determinant $1$, and (2) interchanging two rows or columns of a matrix multiplies its determinant by $-1$.
When using Cramer's rule, there will be no variables whatsoever, rather, you know that the coordinates of the solution to $Ax=b$ are given by $$x_i=\frac{\det{(A\mid i)}}{\det A}$$ where $(A|i)$ is the matrix obtained by replacing $A$'s $i-th$ column with the column vector $b$. I give you an example on how to compute $\det A$, and maybe you can compute the remaining determinants yourself. Recall how how the detereminant behaves:
$({\rm i})$ It remains unchanged if we sum a multiple of a row (column) to another row (column)
$(\rm ii)$ It changes sign if we permute two rows (columns)
$(\rm iii)$ Scalars hop off the determinant.
Thus you may partially triagulate your matrix and operate as follows
$$\begin{align}
\det \left( {\begin{matrix}
1 & 1 & { - 2} & 3 \\
0 & { - 2} & 0 & { - 1} \\
0 & 7 & 3 & 0 \\
0 & 9 & 5 & { - 1} \\
\end{matrix} } \right) &= - \det \left( {\begin{matrix}
1 & 1 & { - 2} & 3 \\
0 & 2 & 0 & 1 \\
0 & 7 & 3 & 0 \\
0 & 9 & 5 & { - 1} \\
\end{matrix} } \right) \cr
\text{cofactor} &= - \det \left( {\begin{matrix}
2 & 0 & 1 \\
7 & 3 & 0 \\
9 & 5 & { - 1} \\
\end{matrix} } \right) \cr
R_2-3R_1\to R_2'\;,R_3-9R'_2\to R_3'\;\; &= - \det \left( {\begin{matrix}
0 & { - 6} & 7 \\
1 & 3 & { - 3} \\
0 & { - 22} & {26} \\
\end{matrix} } \right) \cr
\text{ permute rows } &= \det \left( {\begin{matrix}
1 & 3 & { - 3} \\
0 & { - 6} & 7 \\
0 & { - 22} & {26} \\
\end{matrix} } \right) \cr
\text{ cofactor } &= \det \left( {\begin{matrix}
{ - 6} & 7 \\
{ - 22} & {26} \\
\end{matrix} } \right) = - 2 \end{align} $$
You can also triangulate it, and just calculate $\prod a_{ii}$ which is almost what I did above. Here is W|A's computation.
ADD There is a formula that does not involve cofactors, but it involves $4!=24$ terms, namely $$\det A=\sum_{\sigma \in S_4}\operatorname{sgn}\sigma a_{1\sigma(1)}\cdots a_{n\sigma(n)}$$
Best Answer
Use the following properties:
See 1 and 2 for more properties.