Solve the given linear system by any method

frctl

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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

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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

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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

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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

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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
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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

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Messages
153
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

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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

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Messages
153
Yes, you're right.
 

Otis

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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

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What is this entry supposed to be sorry
 

Otis

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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

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Yes, thank you for helping me.
 

Otis

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

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

😎
 

Otis

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Have you finished with this exercise, yet?

\(\;\)
 

frctl

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Yes
 
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