A = [A-x(1) B-x(2) C-x(3);D-x(4) E-x(5) F-x(6); G-x(7) H-x(8) I-x(9)]
I have to obtain x(1)...x(9)
for det(A) = 0
.
Given a 3x3
matrix A
its determinant is
therefore you need to solve |A| = 0
. For your case we are given
The easiest solution for x
so that |A| = 0
is when
a - x(1) = 0
b - x(2) = 0
c - x(3) = 0
which leads to
x(1) = a
x(2) = b
x(3) = c
so
x = A
is the most trivial solution. There exists an infinite number of solutions to this problem, this is just one. You could choose another solution where
a - x(1) != 0
b - x(2) != 0
c - x(3) != 0
and then you would have to set
ei - fh = 0
di - fg = 0
dh - eg = 0
which would involve simultaneous equations.
I suggest before trying to code up a solution you work through one by hand like I've done here.