# Linear Algebra: Nontrivial and trivial solutions.

#### math951

I understand how to compute augmented matrices, but I am not understanding this specific answer.

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
I understand how to compute augmented matrices, but I am not understanding this specific answer.

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 $0x_1+0x_2+0x_3 = 0$

#### math951

But why do we have to add a third row?

#### math951

Is it because we have 3 columns?

#### math951

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
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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