Although your notation is somewhat non-standard [to fit the formatting capabilities of posting in plain text?], I think I know what you mean. Writing out the first example to just make sure, we have the 3X3 matrix A
A = \(\displaystyle \begin{pmatrix}
1&1&1\\
1&1&1\\
0&0&0
\end{pmatrix}\), the column vector b = \(\displaystyle \begin{pmatrix}
0\\
0\\
0
\end{pmatrix}\), and the variable vector x = \(\displaystyle \begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}\) with the equation
A x = b.
You statement is 'the solution has many solutions for x including a non-trivial solution.' That is a correct statement.
For your second example
111=0
111=0
001=0
it has trivial solution and it has unique solution that is zero?
That statement is not completely correct. It does have the trivial solution, x
1 = x
2 = x
3 = 0, but it is not unique. The first two rows are linearly dependent,
1 * [1 1 1] + (-1) * [1 1 1] = 0
and thus the determinate of the corresponding matrix A is zero in this case. It also has non-trivial solutions, x
2 = -x
1, x
3 = 0, i.e.
x = \(\displaystyle \begin{pmatrix}
x_1\\
-x_1\\
0
\end{pmatrix}\). x1 \(\displaystyle \ne\) 0.
To go over your questions:
1)trivial solution are always have unique solution?
No. Look at the second example above.
2)non trivial solution have always many solution?
No. Look at, in your notation,
1 0 0 = 1
0 1 0 = 2
0 0 5 = 20
The determinate of the system matrix is non-zero and thus has a unique solution. The solution is x
1=1, x
2=2, x
3=4. The solution is non-trivial since at least one of the x's is non-zero (actually all three are non-zero).
3)if a matrix is linearly dependent then it MUST have non trivial solution?
example
111=0
111=0
001=0 here matrix is independent and also have trivial solution
No. If it has a solution, then it must have a non-trivial solution. However, it may be that the system is inconsistent and there is no solution.
To be continued