For a #2\times2# matrix the determinant is
#det[((a,b),(c,d))]=ad-bc#
In a #3\times3# matrix you have to calculate the determinant of three sub-matrices #2\times2#.
We begin selecting the first element
#((\color[red]a,b,c),(d,e,f),(g,h,i))#
then we identify the sub-matrix that is not in the raw or the column of that element
#((\color[red]a,b,c),(d,\color[cyan]e,\color[cyan]f),(g,\color[cyan]h,\color[cyan]i))#
and finally we multiply the element selected (in our case #a#) by the determinant of the #2\times2# matrix
#a(ei-fh)#.
We repeat this process for the elements of the first raw alternating the signs
#((a,\color[red]b,c),(\color[cyan]d,e,\color[cyan]f),(\color[cyan]g,h,\color[cyan]i))#
#-b(di-fg)#
#((a,b,\color[red]c),(\color[cyan]d,\color[cyan]e,f),(\color[cyan]g,\color[cyan]h,i))#
#c(dh-g e)#.
The determinant is then the sum of these three terms:
#a(ei-fh)-b(di-fg)+c(dh-g e)#
We are now ready to calculate our determinant:
#det((5,2,4),(2,-2,-3),(-3,2,5))=#
#5*(-2*5-(-3)*2)#
#-2*(2*5-(-3)*(-3))#
#+4*(2*2-(-2)*(-3))#
#=-20-2-8=-30#.
