- Thread starter frctl
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This is a homogeneous set of equations.Solve the given linear system by any method

2x_{1}+ x_{2}+ 3x_{3}= 0

x_{1}+ 2x_{2}= 0

x_{2}+ x_{3}= 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 ?

What have learned about "homogeneous" equations?

By observation, x

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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).

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:2 1 3 0

1 2 0 0

0 1 1 0

… multiply row1 by 1/2 ?

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 …

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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 \(\displaystyle -2x_1+ x_2- 3x_3= 0\).

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Multiply R2 by (-3) and add to R3.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?

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The last row should be [0 0 3 0]Multiply R2 by (3) and add to R3

1 2 0 0

0 1 1 0

0 0 0 0

therefore

x_{1}+ 2x_{2}= 0

x_{2}+ x_{3}= 0

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

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Bottom row is not correct.…

R3 -> -2*R1 + R3

1 2 0 0

0 1 1 0

0 -3 0 0

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

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.

\(\;\)

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

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.

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

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