Let #A = [a_(ij)] # be an #nxxn# matrix with entries from field F. When finding the determinant of A there are a couple of things we need to do. First, assign each entry a sign from the sign matrix. My linear algebra lecturer called it a "sign chessboard" which has stuck with me.
#((+,-,+,...),(-,+,-,...),(+,-,+,...),(vdots,vdots,vdots,ddots))#
So this means that the sign associated with each entry is given by #(-1)^(i+j)# where #i# is the row of the element and #j# is the column.
Next, we define the cofactor of an entry as the product of the determinant of the #(n-1)xx(n-1)# matrix we obtain by removing the row and column containing that entry and the sign of that entry.
We then obtain the determinant by multiplying each entry in the top row (or column) by it's cofactor and summing these results.
Now that the theory is out the way, let's do the problem.
#A = ((1,4,-2),(3,-1,5),(7,0,2))#
The sign associated with #a_(11)# is +, with #a_(12)# is - and with #a_(13)# is +
We obtain that
#det(A) = color(red)(1)color(blue)([(-1,5),(0,2)]) + color(red)(4)color(blue)((-1)[(3,5),(7,2)] + color(red)((-2))color(blue)([(3,-1),(7,0)])#
Where red denotes the entries from the top row and blue is their respective cofactor.
Using the same method we see that the determinant of a #2xx2# matrix
#det((a,b),(c,d)) = ad-bc#
Hence:
#det(A) = color(red)(1)color(blue)(((-1)*2 - 5*0)) color(red)(-4)color(blue)((3*2-5*7)) color(red)(-2)color(blue)((3*0 - (-1)*7))#
#det(A) = -2 + 116 - 14 = 100#
