If you take the determinant of a #3times3# matrix

#|(a,b,c),(f,g,h),(x,y,z)|#

you take each element in the first row and multiply it by the determinants of #2times2# submatrices which are composed by the elements in the bottom two rows of each column other than the column the particular first row element is from

#|(a,b,c),(f,g,h),(x,y,z)|=a|(g,h),(y,z)|-b|(f,h),(x,z)|+c|(f,g),(x,y)|#

Notice the second term is negatve. The signs alternate.

Then we take the determinants of the submatrices.

#|(a,b,c),(f,g,h),(x,y,z)|=a(gz-hy)-b(fz-hx)+c(fy-gx)#

In this case our matrix is

#|(2,-1,3),(3,0,-2),(1,-3,0)|#

Then

#|(2,-1,3),(3,0,-2),(1,-3,0)|=2|(0,-2),(-3,0)|-(-1)|(3,-2),(1,0)|+3|(3,0),(1,-3)|#

#=2(0-6)+(0-(-2))+3(-9-0)#

#=2(-6)+2+3(-9)=-12+2-27#

#=-12-27+2=-39+2=-37#