Tolong bantu jawab sama caranya

p^-1 = 1/det P x Adjoin P
P^-1 = 1/(1.3-1.2) x [3...-2]
……………………........ [-1....1]

p^-1 = 1/1 x [3...-2]
.....................[-1....1]
p^-1 = [3....-2]
............[-1......1]

2p^-1 = [6...-4]
..............[-2...2]
[x .....y] = [ 6...-4]
[-z.....z]. [-2...2]

jadi x = 6 , y = -4, z = 2

x+y = 6 + (-4) = 2