x 1-x2-2x4=3

x 1-2x2+4x3-3x4=λ

Perform line transformation (and only line transformation)

Switch lines first.

x 1-x2-2x4=3

x 1-2x2+4x3-3x4=λ

2x 1-3x2+4x3-5x4= 1

The augmented matrix is: R(A|b)

1 - 1 0 -2 3

1 -2 4 -3 λ

2 -3 4 -5 1

Then, line 1 is multiplied by (-1) respectively, and line 2 and line 3 are added with (-2).

1 - 1 0 -2 3

0 - 1 4 - 1 λ-3

0 - 1 4 - 1 -5

The second line is multiplied by (-1) and added to the third line.

1 - 1 0 -2 3

0 - 1 4 - 1 λ-3

0 0 0 0 -2-λ

To make the equation have a solution:

R(A)=R(A|b)

So -2-λ=0.

λ=-2

At this time, it becomes:

1 - 1 0 -2 3

0 - 1 4 - 1 -5

0 0 0 0 0

Line 2 is multiplied by (-1) and added to line 1, and then line 2 is multiplied by (-1).

1 0 -4 - 1 8

0 1 -4 1 5

0 0 0 0 0

Then we know that there is a special solution of (8,5,0,0) t (where t stands for transposition).

The general solution is the general solution corresponding to the homogeneous equation corresponding to the nonhomogeneous linear equations:

(4，4， 1，0)T

( 1，- 1，0， 1)T

So the solution of the equation is:

k 1( 4，4， 1，0)T +k2( 1，- 1，0， 1)T+(8，5，0，0)T