19: Let the coefficient matrix be A and the matrix on the right of the equal sign be B, which can be written as the inverse matrix of AX=B, with both sides multiplied by A on the left, C, CAX=CB, CA=E (identity matrix) and X=CB. To find the inverse matrix, you can spell a identity matrix in front of A, transform the matrix on the right into identity matrix, and the matrix on the left becomes the inverse matrix:

1 0 0 1 - 1 1

0 1 0 2 3 0

0 0 1 0 2 - 1

Add the last line to 1 line:

1 0 1 1 1 0

0 1 0 2 3 0

0 0 1 0 2 - 1

Line 2- Line 1 x2

1 0 1 1 1 0

-2 1 -2 0 1 0

0 0 1 0 2 - 1

65438 Line +0-2 Line, 3 Line -2 Line x2

3 - 1 3 1 0 0

-2 1 -2 0 1 0

4 -2 5 0 0 - 1

The last line x(- 1)

3 - 1 3 1 0 0

-2 1 -2 0 1 0

-4 2 -5 0 0 1

The reciprocal of a:

3 - 1 3

-2 1 -2

-4 2 -5

A times the reciprocal of b.

1 1 12 -3 27 26

-6 -8 1 - 18 - 17

- 15 -2 1 9 -38 -35

There are n unknowns. The system of n-order homogeneous linear equations AX=0 has only zero solution, and k is an arbitrary positive integer. It is proved that KX = 0 also has zero solution.

The equation system AX=0 has only zero solution, then the determinant of A ||||≠ 0 has an inverse A(- 1), so that A(- 1)A=E (identity matrix). Multiply both sides by A(- 1):

A(- 1)A^kX=A(- 1)0=0

a^(k- 1)x=0；

Repeat the above steps, and finally get:

AX=0

There is only zero solution.