Rozwiąż układ Zbyszek:

⎧	(1−i)x−(2+i)y = i
⎨
⎩	3ix+(2−i)y=2+3i

bardzo proszę o pomoc

Basia: jak każdy układ (1−i)x = i+(2+i)y

	i+(2+i)y		[i+(2+i)y](1+i)		i + i2 + (2+i)(1+i)y
x =		=		=		=
	1−i		1−i2		2

i−1+(2+3i−1)y		i−1+(1+3i)y
	=
2		2

	i−1+(1+3i)y
3i*		+ (2−i)y = 3i /*2
	2

3i*[i−1+(1+3i)y] + 2(2−i)y = 6i 3i² − 3i + (3i+9i²+4−2i)y = 6i (i − 5)y = 9i+3

	9i+3		(9i+3)(i+5)		9i2+45i+3i+15
y =		=		=		=
	i−5		i2−25		−26

6 + 48i		−3−24i
	=
−26		13

podstaw teraz za y i wylicz x

Mila: W=| (1−i) (−2−i) | |3i (2−i) | W=(1−i)*(2−i)−3i*(−2−i)=3i−2 W_x=|i −2−i | |2+3i 2−i | W_x=i(2−i)−(−2−i)(2+3i)=2i−i²−(−4−6i−2i−3i²)=2i+1−(−4−8i+3)=10i+2 W_y=|1−i i | | 3i 2+3i| W_y=(1−i)(2+3i)−3i²=2+3i−2i−3i²+3=i+8

	Wx		2+10i
x=		=		=... dokończ
	W		−2+3i

	Wy		8+i
y=		=		=... dokończ
	W		−2+3i