Cofactor expansion to check if matrices is invertible. I have a question regarding a co-factor expansion question. I understand that an easy way to check if a matrix is invertible is to do co-factor expansion and if $A \ne 0$ then it's invertible. I'm familiar with the basic co-factor expansion with $3 \times 3$ matrices although I'm unfamiliar with $4\times 4$ matrices.
Could someone please give me a hint, or tell me how to go about checking if this matrix is invertible? 
Q: Use the inversion algorithm to find the inverse of the matrix (if the inverse exists)
$$\begin{bmatrix}0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & -1 & 3 & 0\\ 2 & 1 & 5 & -3 \end{bmatrix}$$
 A: To check if matrices are invertible, you need to check the determinant is non-zero: 
To find the determinant of this matrix we look for the row or column with the most zeros and do a Laplace development on that row or column.
The first row contains the most zeros so we Laplace develop that row:
$$\det=\begin{vmatrix}0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & -1 & 3 & 0\\ 2 & 1 & 5 & -3 \end{vmatrix}$$  
$$=0\times\begin{vmatrix} 0 & 0 & 1 \\  -1 & 3 & 0 \\ 1 & 5 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 1 \\  0 & 3 & 0 \\ 2 & 5 & -3 \end{vmatrix}+2\times\begin{vmatrix} 1 & 0 & 1 \\  0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 0 \\  0 & -1 & 3 \\ 2 & 1 & 5 \end{vmatrix}$$$$=2\times\begin{vmatrix} 1 & 0 & 1 \\  0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}$$ $$=2\left( 1\times\begin{vmatrix}-1 & 0 \\ 1 & -3 \\ \end{vmatrix}+0\times\begin{vmatrix}0 & 0 \\ 2 & -3 \\ \end{vmatrix}+ 1\times\begin{vmatrix}0 & -1 \\ 2 & 1 \\ \end{vmatrix}\right)$$$$=2\times \left(1\times(3-0)+1\times(0--2)\right)= 2(3+2)=10 \ne 0$$ So your matrix is invertible (or has an inverse). By the way, when I say 'Laplace development', I think that's what you meant by co-factor expansion. They are not the same thing because the matrix of co-factors in something different. Does this help? Or is there something else I can do? 
