Здравствуйте, Чаркин Иван Александрович.

....|2 1 1|___|3|_________________| 1|
A=|1 3 1| b=|0| точное решение x=|-1|
....|1 1 2|___|4|_________________| 2|

LU-разложение
A=L*U, L*y=b, U*x=y

U₁₁=a₁₁=2
U₁₂=a₁₂=1
U₁₃=a₁₃=1
L₁₁=a₁₁/u₁₁=1
L₂₁=a₂₁/u₁₁=1/2=0.5
L₃₁=a₃₁/u₁₁=1/2=0.5
U₂₂=a₂₂-l₂₁*u₁₂=3-1*1/2=5/2=2.5
L₂₂=(a₂₂-l₂₁*u₁₂)/u₂₂=(3-1*1/2)/(5/2)=1
U₂₁=a₂₁-l₂₁*u₁₁=1-2*1/2=0
L₁₂=(a₁₂-l₁₂*u₂₂)/u₁₁=(1-2*1/2)/2=0
U₂₃=a₂₃-l₂₁*u₁₃=1-1*1/2=1/2=0.5
L₃₂=(a₃₂-l₃₁*u₁₂)/u₂₂=(1-1*1/2)/(5/2)=1/5=0.2
U₃₁=a₃₁-l₃₁*u₁₁-l₃₂*u₂₁=1-2*1/2-0*1/5=0
U₃₃=a₃₃-l₃₁*u₁₃-l₃₂*u₂₃=2-1*1/2-(1/5)*(1/2)=14/10=7/5=1.4
U₃₂=a₃₂-l₃₁*u₁₂-l₃₂*u₂₂=1-1*1/2-(1/5)*(5/2)=0
L₁₃=(a₁₃-l₁₁*u₁₃-l₁₂*u₂₃)/u₃₃=(1-1*1-0*1/2)/(7/5)=0
L₂₃=(a₂₃-l₂₁*u₁₃-l₂₂*u₂₃)/u₃₃=(1-1*1/2-1*1/2)/(7/5)=0
L₃₃=(a₃₃-l₃₁*u₁₃-l₃₂*u₂₃)/u₃₃=(2-1*1/2-(1/2)*(1/5))/(7/5)=1

L*y=b

1*У₁=3 -> y₁=3
(1/2)*y₁+1*y₂=0 -> y₂=-3/2 =*1.5
(1/2)*y₁+(1/5)*y₂+1*y₃=4 -> y₃=14/5=2.8

U*x=y
(7/5)*x3=14/5 -> x3=2
(5/2)*x2+(1/2)*x3=-3/2 -> x2=-1
2*x1+1*x2+1*x3=3 -> x1=1

Метод Холецкого

A=L*LT
L*y=b, LT*x=y

L₁₁=sqrt(a₁₁)=sqrt(2)=1.414
L₂₁=a₂₁/l₁₁=1.414/2=0.707
L₂₂=sqrt(a₂₂-l₂₁²)=1.581
L₃₁=a₃₁/l₁₁=1/1.414=0.707
L₃₂=(a₃₂-l₃₁*l₂₁)/l₂₂=(1-0.707*0.707)/1.581=0.316
L₁₂=l₁₃=l₂₃=0

L*y=b
1.414*y₁=3 -> y₁=3/1.414=2.121
0.707*y₁+1.581*y₂=0 -> y₂=-0.949
0.707*y₁+0.316*y₂+1.183*y₃=4 -> y₃=2.366

LT*x=y
1.183*x₃=2.366 -> x₃=2.366/1.183=2
1.581*x₂+0.316*x₃=-0.949 -> x₂=(-0.949-0.316*2)/1.581=-1
1.414*x₁+0.707*x₂+0.707*x₃=2.121 -> x₁=(2.121-0.707*(-1)-0.707*2)/1.414=1

Ответ: x[sub]1[/sub]=1 ; x[sub]2[/sub]= -1 ; x[sub]3[/sub]= 2