My Math Forum non trivial solution

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 December 22nd, 2014, 01:38 AM #1 Member   Joined: Dec 2014 From: India Posts: 61 Thanks: 0 non trivial solution 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
December 22nd, 2014, 01:46 PM   #2
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Quote:
 Originally Posted by 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 should have non trivial solution.am i right? Thank you
2) should be: determinant not zero.

It has one solution: x=y=z=0.

 December 22nd, 2014, 02:05 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra I think there can be nontrivial solutions when the determinant is zero, but there are not unique solutions in this case, having at least one free variable.
 December 22nd, 2014, 08:12 PM #4 Member   Joined: Dec 2014 From: India Posts: 61 Thanks: 0 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
 December 22nd, 2014, 09:10 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra The trivial solution is $x = y = z = 0$. A non-trivial solution is any solution that is not $x = y = z = 0$. If there is only one (non-trivial) solution, it is unique. If there is more than one (non-trivial) solution, they are not unique. Consider $$x + y + z = 0 \\ 2x + 2y + 2z = 0 \\ x + y = 0$$ This has an infinite number of non-trivial solutions $(a, -a, 0)$ where $a$ can be any real number. This happens because the three equations are not linearly independent. In particular, the first and second equations are equal. This means that in effect we have only two constraints on the three variables. The determinant is zero because the equations are linearly dependant. In the case that the three equations are linearly independent, the only solution is trivial. This is because the matrix now has an inverse. $$A\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} \implies A^{-1}A\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}x \\ y \\ z\end{pmatrix} = A^{-1}\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$$ Thanks from Bhuvaneshnick
December 22nd, 2014, 09:52 PM   #6
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Quote:
 The determinant is zero because the equations are linearly dependant.
if determinant is zero the equations are linearly dependent and ALSO HAVE NON TRIVIAL SOLUTION?

Quote:
 In the case that the three equations are linearly independent, the only solution is trivial. This is because the matrix now has an inverse.
what does that mean matrix has inverse can you explain me with other example.please

 December 22nd, 2014, 10:19 PM #7 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 The trivial solution is the unique solution where every variable equals zero. If you have infinitely many solutions, you have non-trivial solutions. One thing you'll need to be careful of is how many equations you have and how many variables you have. $\displaystyle x + y + z = 0 \ \ \ (1) \\\;\\ 2x + 2y + 2z = 0\ \Longrightarrow\ 2(x+y+z)=0|_{:2}\Longrightarrow\ x+y+z=0\ \ \ (2) \\\;\\ x + y = 0\ \ \ (3) \\\;\\ (1), (2),\ (3) \Longrightarrow\ There \ is\ \ 2 \ equations\ and \ 3\ variables. \\\;\\ x + y + z = 0 \\\;\\ x+y=0$ We note $\displaystyle x = a$ and system solutions are $\displaystyle (a, -a, 0)$, $\displaystyle a \in \mathbb{R}$ For a = 0, we obtain the trivial solution (0, 0, 0). Thanks from Bhuvaneshnick
 December 22nd, 2014, 11:15 PM #8 Member   Joined: Dec 2014 From: India Posts: 61 Thanks: 0 thanks for all i under stand now.please finally answer this one 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 non 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
December 23rd, 2014, 06:38 AM   #9
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Quote:
 Originally Posted by Bhuvaneshnick it has non trivial solution with many solution,is that right?
Yes.
Quote:
 Originally Posted by Bhuvaneshnick it has non trivial solution and it has unique solution that is zero?
Zero is always a solution when there are zeroes on the right.
Quote:
 Originally Posted by Bhuvaneshnick 1)trivial solution are always have unique solution?
The trivial solution is always a solution,, but it is not always the unique solution.
Quote:
 Originally Posted by Bhuvaneshnick 2)non trivial solution have always many solution?
Correct.
Quote:
 Originally Posted by Bhuvaneshnick 3)if a matrix is linearly dependent then it MUST have non trivial solution?
Correct .
Quote:
 Originally Posted by Bhuvaneshnick 5)if the determinant of co-efficient is zero then it has non trivial solution?
Correct.

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