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.
