Question

How do you find the determinant of `((0, 1, 2, 3), (0, 0, 2, 0), (3, 10, 12, 20), (0, 0, 0, 4))`?

Answer 1

given by

`-2det((0, 1, 3),(3,10,20),(0,0,4))`

and

`det((0, 1, 3),(3,10,20),(0,0,4))`

is given by

`4det((0,1),(3,10))`

and

`det((0,1),(3,10))`

is given by

`0*10-3*1 = -3`

In fact you can develop on what line or column you want, find one which have most `0`

Then remember formula

`det(A) = sum_(n=0)^(n)(-1)^(i+(j+n))a_(i(j+n))det(A_(i(j+n)))`

Complex ? not really

here you develop the determinant in line because `i` is fixed (and `=2`, you just have to read the number which is `a_(ij)` and to have `A_(ij)` just hide with your finger the column and the line which contain `a_(ij)`

here it's simple, `a_(21) = a_(22) = a_(24) = 0` so you just have

`det(A) = (-1)^(2+3)a_(23)det(A_(23))`

repeat that until you have `2xx2` matrix like i did