non trivial solution

[bhuvaneshnick]

I was thinking that non trivial solution occur in following case say whether i am right or wrong
1) when number of equation = number of variables
2)determinant of co-efficient is zero


According to first
2x+3y=0
5x+5y=0
4x+5y+7z=0
It shoud have non trivial solution.am i right?
Thank you

 

[HallsofIvy]

Yes, that is an "evident" solution but the question is whether there are or are not other solutions.

bhuvaneshnick, it is true that "If the number of equations is equal to the number of variables and the determinant of the coefficient matrix is non-zero, then there is a unique solution"- and it is "evident" that x= y= z= 0 is a solution. If the determinant is 0 then there are an infinite number of solutions so "non-trivial" solutions.

However, you are mistaken in thinking that the determinant of the coefficient matrix is 0!

The determinant is \(\displaystyle \left|\begin{array}{ccc}2 & 3 & 0 \\ 5 & 5 & 0 \\ 4 & 5 & 7 \end{array}\right|= 7\left|\begin{array}{cc} 2 & 3 \\ 5 & 6\end{array}\right|= 7(-3)= -21\).

The only solution to this system of equations is x= y= z= 0.

 

[Ishuda]

I was thinking that non trivial solution occur in following case say whether i am right or wrong
1) when number of equation = number of variables
2)determinant of co-efficient is zero
...

Did you mean
2)determinant of co-efficient matrix is non-zero?
If so, it may be that you do not mean a non trivial solution occurs but rather that a unique solution occurs. In the example you gave the determinate of the coefficient matrix is non-zero and there are as many equations as variable. The unique solution is x=y=z=0 as Denis pointed out.

In fact, to tie this in slightly to your other post, there is always a non-trivial solution for the homogeneous under determined system (more variables than equations), see
http://en.wikipedia.org/wiki/Underdetermined_system
and, for general inverses of matrices,
http://en.wikipedia.org/wiki/Generalized_inverse
for example.

 

[bhuvaneshnick]

i have few doubts in the above post.but before that i would like to know the difference between among trivial, non trivial ,unique solution.it may be please answer Thank you

 

[Ishuda]

i have few doubts in the above post.but before that i would like to know the difference between among trivial, non trivial ,unique solution.it may be please answer Thank you

trivial solution: all variable values equal zero is a solution.

non-trivial solution: at least one of the variables in the solution is non zero.

unique solution: If a is a solution and b is a solution then a = b. Note that a and b could be scalars, row vectors, matrices, etc.

 

[bhuvaneshnick]

111=0
111=0
000=0
it has non trivial solution with many solution,is that right?
111=0
111=0
001=0
it has trivial solution and it has unique solution that is zero?
i hope the above are correct let me check my following assumption are correct or not
1)trivial solution are always have unique solution?
2)non trivial solution have always many solution?
if the above is not correct give me example for trivial with many solution and non trivial with unique solution
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
4)if a matrix is linearly dependent then it MUST have non trivial solution?
example
111=0
111=0
000=0 here matrix is dependent it must have non trivial solution
5)if the determinant of co-efficient is zero then it has non trivial solution?
6)if the determinant have of co-efficient is non zero then it have trivial solution ?
Thank you in advance

 

[Ishuda]

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, x₁ = x₂ = x₃ = 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, x₂ = -x₁, x₃ = 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 x₁=1, x₂=2, x₃=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

 

[Ishuda]

I don't know why but it seems as though I can't post anything 'formatted' on here now. For example, if I try to post the below without the 'code' tags, it is not allowed.

First, as a matter of convention, what we are discussing here is called, variously, a system of linear equations,
 matrix equations, vector equations, etc.  A matrix is singular (has determinate zero) or non-singular (has a non-zero 
determinate) and the (row and column) vectors are linearly dependent/independent.  Thus your statements in (3)  'if 
a matrix is linearly dependent' and 'matrix is independent' should be ' matrix is singular' and 'matrix is non-singular'. 
 BTW: Your statement 'here matrix is independent and also have trivial solution' is only partially correct.  See the 
discussion of your example 2 above.

To continue with question (3):  However, it may be that the system is inconsistent and there is no solution. As an 
example of an inconsistent system of equations, consider
1 0 0 = 0
1 0 0 = 1
0 0 1 = 1
Row 1 and 2 are linearly dependent, i.e.
1 * [1 0 0] + (-1) * [1 0 0] = 0,
the determinate of the matrix of the system is zero and there is no solution.  However, if you change the wording 
slightly to 'If the determinate of the system is zero for a homogenous system (b is the zero column vector), the 
system has a non-trivial solution.' then that is a true statement.

[quote=bhuvaneshnick]4)if a matrix is linearly dependent then it MUST have non trivial solution?... [/quote]
No.  See discussion of question (3) above.

[quote=bhuvaneshnick] 5)if the determinant of co-efficient matrix is zero then it has non trivial solution?[/quote]
No.  See discussion of question (3) above.

[quote=bhuvaneshnick] 6)if the determinant have of co-efficient matrix is non zero then it have trivial solution ?[/quote]
No.  If the determinate of the co-efficient matrix is non-zero, the it has a unique solution which depend on the 
column vector b.  If the system is homogenous (b is the zero column vector), the unique solution is the trivial 
solution.