Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:
$$A = \left(
        \begin{matrix}
        5&-7&2&2\\
        0&3&0&-4\\
        -5&-8&0&3\\
        0&5&0&-6\\
        \end{matrix} \right)
$$
I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
        \begin{pmatrix}
        3 & 0 & -4 \\
        -8 & 0 & 3 \\
        5 & 0 & -6 \\
        \end{pmatrix} 
$$
M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:
3 times $$
        \begin{pmatrix}
        0 & 3 \\
        0 & -6  \\
        \end{pmatrix}
$$ 
giving $3(0-0)= 0$ 
then:
0 times $$
        \begin{pmatrix}
        -8 & 3\\
         5 & -6\\
        \end{pmatrix}
$$
giving 0(48-15)=0
Then:
4 times
$$
        \begin{pmatrix}
        -8 & 0 \\
         5 & 0 \\
        \end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$ 
So det M1 $= 0(1) = 0$
M2--> M(1,2)---> $-1^1+2= -1^3 = -1$
$$
        \begin{pmatrix}
        0 & 0 & -4 \\
        -5 & 0 & 3 \\
        0 & 0 & -6 \\
        \end{pmatrix}
$$
o* 
$$
        \begin{pmatrix}
        0 & 3 \\
        0 & -6 \\
        \end{pmatrix}
$$
giving $0(0-0)=0$
obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally 
4 times
$$
        \begin{pmatrix}
        -8 & 0 \\
        5 & 0 \\
       \end{pmatrix}
$$
giving 4(0-0)= 0 
So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3
M3 --> $-1^4 = 1$
$$
        \begin{pmatrix}
        0 & 3 & -4 \\
        -5 & -8 & 3 \\
        0 & 5 & -6 \\
        \end{pmatrix}
$$
for the determinant:
0 times
$$
        \begin{pmatrix}
        -8 & 3 \\
         5 & -6 \\
        \end{pmatrix}
$$
which gives $0(48-15)=0$
-3 times
$$
        \begin{pmatrix}
        -5& 3 \\
        0 & -6 \\  
        \end{pmatrix}
$$
which gives $-3(30-0)= -90$
it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4. 
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving 
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases? 

Write out the determinant. Alongside write the first  3 columns"Downward diagonal multiplication" gives ( O + O + 2OO + O ).  "Upward diagonal multiplication" gives -(O +  O + O + 18O) . det = 2O.
