Linear Algebra: Nontrivial and trivial solutions.

Jul 2015
697
25
United States
I understand how to compute augmented matrices, but I am not understanding this specific answer.

Please look here:

Linear Algebra and Its Applications (9780321982384), Pg. 48, Ex. 4 :: Homework Help and Answers :: Slader

I do not understand the answer as to why that is like that. (I know trivial means where the only solution to Ax=0, is where x=0, and non-trivial exists where Ax=0, and x does not = 0.

how is x3 free?

Free implies where there are all 0's in the row does it not?

My buddy told me that the x3 is free because "the third row is missing." That does not make any sense to me, this problem was given 2 equations with 3 variables in each equation, denoting only 2 ROWS.
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
I understand how to compute augmented matrices, but I am not understanding this specific answer.

Please look here:

Linear Algebra and Its Applications (9780321982384), Pg. 48, Ex. 4 :: Homework Help and Answers :: Slader

I do not understand the answer as to why that is like that. (I know trivial means where the only solution to Ax=0, is where x=0, and non-trivial exists where Ax=0, and x does not = 0.

how is x3 free?

Free implies where there are all 0's in the row does it not?

My buddy told me that the x3 is free because "the third row is missing." That does not make any sense to me, this problem was given 2 equations with 3 variables in each equation, denoting only 2 ROWS.
Add a third row $0x_1+0x_2+0x_3 = 0$
 
Jul 2015
697
25
United States
But why do we have to add a third row?
 
Jul 2015
697
25
United States
Is it because we have 3 columns?
 
Jul 2015
697
25
United States
So I am assuming the amount of variables we have dictates how many rows we should have. I.E. 3 variables in equation, I should assume there are 3 equations needed in total, therefore 3 rows.
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
So I am assuming the amount of variables we have dictates how many rows we should have. I.E. 3 variables in equation, I should assume there are 3 equations needed in total, therefore 3 rows.
We can write the same problem like this:

$-5x_1+7x_2+9x_3 = 0$
$x_1-2x_2+6x_3=0$

For the second equation, we can multiply both sides by 5:
$5x_1-10x_2+30x_3=0$

Adding this to the first equation gives:

$0x_1-3x_2+39x_3=0$

Solving for $x_2$ we get: $x_2=13x_3$. Now we can plug this back into the second equation:

$x_1-2x_2+6x_3=0$
$x_1-2(13x_3)+6x_3=0$
$x_1=20x_3$

So, $x_3$ is a free variable, and we can express both $x_1$ and $x_2$ in terms of $x_3$. Namely, $x_1=20x_3$ and $x_2=13x_3$. The matrix operations simply help you do it.
 
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