cofactor expansion

This is a recursive method, simple to do by hand for small matrices but inefficient for larger matrices:

$$\det (A)=\sum _{j=1}^n(-1{)}^{i+j}{a}_{i,j}{m}_{i,j}=\sum _{i=1}^n{a}_{i,j}{C}_{i,j}$$

The minor $m_{i,j}$ is the determinant of a $(n-1)\times (n-1)$ sub-matrix obtained by omitting the row $i$ and column $j$.

Link to original

$C_{ij}:=(-1{)}^{i+j}\cdot m_{i,j}$ is called the cofactor.

For example:

$$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 5& 0\\ 7& 8& 9\end{array}\right]$$

First choose a row ($i$), preferrably one with many $a_{i,j}$ being zero. $i=2$ in this case.
Create $n$ minor matrices by crossing out row $i$ and columns $1-n$:

\begin{bmatrix} 
\square & 2 & 3 \\ 
\square & \square & \square \\  
\square & 8 & 9
\end{bmatrix}
\rightarrow 
\begin{vmatrix} 2 & 3 \\ 8 & 9 \end{vmatrix}
= m_{2,1}

Link to original

Multiply the determinant of each minor by the corresponding element of the original matrix $a_{i,j}$:

$$-0\cdot \left|\begin{array}{cc}2& 3\\ 8& 9\end{array}\right|+5\cdot \left|\begin{array}{cc}1& 3\\ 7& 9\end{array}\right|-0\cdot \left|\begin{array}{cc}1& 2\\ 7& 8\end{array}\right|=-60=\det (A)$$

Now we see the benefit of choosing the row with the most zeros.

Should there be more zeros in a col rather than a row, you can simply flip $i$ for $j$ (make the chosen col static and loop over the rows).

If you have bigger than $3\times 3$ matrices, apply the process recursively, i.e. split the matrices until you get to $2\times 2$ matrices.