# Solve the given linear system by any method

#### frctl

##### Junior Member
Solve the given linear system by any method

2x1 + x2 + 3x3 = 0
x1 + 2x2 = 0
x2 + x3 = 0

Augmented matrix
2 1 3 0
1 2 0 0
0 1 1 0

Should I begin with a scale operation,
multiply row1 by 1/2 ?

#### Subhotosh Khan

##### Super Moderator
Staff member
Solve the given linear system by any method

2x1 + x2 + 3x3 = 0
x1 + 2x2 = 0
x2 + x3 = 0

Augmented matrix
2 1 3 0
1 2 0 0
0 1 1 0

Should I begin with a scale operation,
multiply row1 by 1/2 ?
This is a homogeneous set of equations.

What have learned about "homogeneous" equations?

By observation, x1 = x2 = x3 = 0

#### Otis

##### Senior Member
Hi frctl. The RREF form of that augmented matrix looks like this:

1 0 0 0
0 1 0 0
0 0 1 0

That shows the trivial solution, posted by Subhotosh. You can check your textbook or google, for more information about homogeneous systems (they have either one solution or infinite solutions, so they are always consistent).

2 1 3 0
1 2 0 0
0 1 1 0
… multiply row1 by 1/2 ?
There's a better way to start, if you would like to get RREF form. Swapping row positions is a valid row operation. Note that moving row1 to the bottom would give us:

1 2 0 0
0 1 1 0
2 1 3 0

Look. Now there are pivots (leading ones) in the first two rows, and we didn't have to multiply or add anything to get them.

Two swaps will move row1 to the bottom:
R1 <-> R2 first, and then R2 <-> R3

Next, finish getting zeros below the pivot in col1:
R3 -> -2*R1 + R3

Continue …

#### HallsofIvy

##### Elite Member
Again jumping directly to matrices! Much simpler:'

The second equation says that $$\displaystyle x_1= -2x_2$$ and the third equation says that $$\displaystyle x_3= -x_2$$. Putting those into the first equation, $$\displaystyle 2x_1+ x_2+ 3x_3= 2(-2x_2)+ x_2+ 3(-x_2)= -6x_2= 0$$ so $$\displaystyle x_1= x_2=x_3= 0$$. That's no surprise. It would have been more interesting if the first equation had been $$\displaystyle -2x_1+ x_2- 3x_3= 0$$.

#### frctl

##### Junior Member
R1 <-> R2 first, and then R2 <-> R3
1 2 0 0
0 1 1 0
2 1 3 0
R3 -> -2*R1 + R3
1 2 0 0
0 1 1 0
0 -3 0 0

How can I eliminate the -3 next?

#### Subhotosh Khan

##### Super Moderator
Staff member
R1 <-> R2 first, and then R2 <-> R3
1 2 0 0
0 1 1 0
2 1 3 0
R3 -> -2*R1 + R3
1 2 0 0
0 1 1 0
0 -3 0 0

How can I eliminate the -3 next?
Multiply R2 by (-3) and add to R3.

#### frctl

##### Junior Member
Multiply R2 by (3) and add to R3
1 2 0 0
0 1 1 0
0 0 0 0
therefore
x1 + 2x2 = 0
x2 + x3 = 0

I multiplied by 3 instead of -3 otherwise I obtain -6

#### Subhotosh Khan

##### Super Moderator
Staff member
Multiply R2 by (3) and add to R3
1 2 0 0
0 1 1 0
0 0 0 0
therefore
x1 + 2x2 = 0
x2 + x3 = 0

I multiplied by 3 instead of -3 otherwise I obtain -6
The last row should be [0 0 3 0]

#### frctl

##### Junior Member
Yes, you're right.

#### Otis

##### Senior Member

R3 -> -2*R1 + R3
1 2 0 0
0 1 1 0
0 -3 0 0
Bottom row is not correct.

How can I eliminate the -3 next?
Change -3 into 0 by adding 3. There is no other way.

Where does the 3 come from? It comes from multiplying the pivot by 3. (There is a pivot in the same column as -3.) In other words, multiply the row which contains that pivot by 3, and add the results to the row with -3.

In the second column, the arithmetic is 3(1) - 3 = 0.

$$\;$$

#### frctl

##### Junior Member
What is this entry supposed to be sorry

#### Otis

##### Senior Member
The last row should be [0 0 3 0]
Subhotosh, frctl made a mistake in the bottom row, earlier (post #5).

The last row in post #7 ought to be [0 0 6 0].

I think frct continues posting multiple errors in threads because they're not proofreading their posts or double-checking their arithmetic.

#### frctl

##### Junior Member
Yes, thank you for helping me.

#### Otis

##### Senior Member
What is this entry supposed to be sorry
You tell us!

That is, do the arithmetic correctly, and then you'll know.

#### Otis

##### Senior Member
Have you finished with this exercise, yet?

$$\;$$

Yes