This gives us:
$$\text{det}(A) = 3 \cdot \text{det}\left(\begin{array}{rrr}
6 & 11 & 12 \\
1 & -1 & 2 \\
5 & 2 & 10
\end{array}\right) - 0 + 7 \cdot \text{det}\left(\begin{array}{rrr}
2 & 6 & 12 \\
4 & 1 & 2 \\
1 & 5 & 10
\end{array}\right) - 0$$
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