Using inverse matrix to solve linear equations, the first line X1 + 2x2 + 3x3 = 1, the second line 2x1 + 2x2 + 5x3 = 2, and the third line 3x1 + 5x2 + X3 = 3

Using inverse matrix to solve linear equations, the first line X1 + 2x2 + 3x3 = 1, the second line 2x1 + 2x2 + 5x3 = 2, and the third line 3x1 + 5x2 + X3 = 3

AX = B => X = A⁻¹B
┏[ 1]━[ 2]━[ 3]┓┏[ x]┓ ┏[ 1]┓
┃[ 2]━[ 2]━[ 5]┃┃[ y]┃=┃[ 2]┃
┗[ 3]━[ 5]━[ 1]┛┗[ z]┛ ┗[ 3]┛
┃[ 1]━[ 2]━[ 3]┃
Determinant: ┃ [2] ━ [2] ━ [5] ┃ = 15
┃[ 3]━[ 5]━[ 1]┃
┏[ 1]━[ 2]━[ 3]┓-1 ┏[-23]━[ 13]━[ 4]┓
Inverse matrix: ┃ [2] ━ [2] ━ [5] ┃ = (1 / 15) ┃ [13] ━ [- 8] ━ [1] ┃
┗[ 3]━[ 5]━[ 1]┛ ┗[ 4]━[ 1]━[ -2]┛
┏[ x ]┓ ┏[ 1]━[ 2]━[ 3]┓-1┏[ 1]┓ ┏[-23]━[ 13]━[ 4]┓┏[ 1]┓
┃[ y ]┃= ┃[ 2]━[ 2]━[ 5]┃ ┃[ 2]┃= (1/15)┃[ 13]━[ -8]━[ 1]┃┃[ 2]┃
┗[ z ]┛ ┗[ 3]━[ 5]━[ 1]┛ ┗[ 3]┛ ┗[ 4]━[ 1]━[ -2]┛┗[ 3]┛
┏[15]┓ ┏[ 1]┓
= (1/15)┃[ 0]┃= ┃[ 0]┃
┗[ 0]┛ ┗[ 0]┛
The solution is x = 1, y = z = 0

The following linear equations X1 + 2x2 + X3 = 3 - 2x1 + x2-x3 = - 3 X1 + 4x2 + 2x3 = - 5 are solved by elementary transformation of matrix

┏ 1 2 1 | 3 ┓
┃-2 1 -1 | -3 ┃
| 1 4 2 | - 5 | → 2 × the first line added to the second line. - 1 × the first line added to the third line. (keep the first line) →
┏ 1 2 1 | 3 ┓
┃ 0 5 1 | 3 ┃
| 0 2 1 | - 8 | → keep the third line. →
┏ 1 0 0 | 11┓
┃0 1 -1 | 19┃
| 0 2 1 | - 8 | → keep the second line →
┏ 1 0 0 | 11┓
┃0 1 -1 | 19┃
┗0 0 3 | -46┛→→
┏ 1 0 0 | 11┓
┃0 1 0 | 11/3┃
┗0 0 1 | -46/3┛
x1＝11.x2＝11/3 x3＝-46/3.

Find the proper solution of linear equations in the first line X1 + x2 + 4x3 = 4, the second line - X1 + 4x2 + X3 = 16, and the third line x1-x2 + 2x3 = - 4

1 1 4 4 1 1 4 4 1 1 4 4 1 0 3 0 x1+ 3x3 = 0
-1 4 1 16 0 5 5 20 0 1 1 4 0 1 1 4 x2+x3 = 4
1 -1 2 -4 0 -2 -2 -8 0 1 1 4 0 0 0 0
After the elimination transformation, there are only two equations: three unknowns and two equations
x1=-3x3 (1)
x2=4-x3 (2)
x3=x3 (3)
(1) (2) (3) are the solutions of the original linear equations

Kramer's rule for solving equations
|2 -1 3 2 |
|3 -3 3 2 |
|3 -1 -1 2 |
|3 -1 3 -1|
right
six
five
three
four
The way I know it is to work it out is different from the answer,

D=D1=D2=D3=D4=-70
When doing row exchange and row addition, you can calculate several of them together, which is equivalent to the elementary transformation of the original equations, and does not affect the value of the determinant. When calculating the determinant, you can be more careful