# Thread: Linearly Independent (Matrix), help!

1. ## Linearly Independent (Matrix), help!

Linearly independent (MAtrix)

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The rows of the matrix:
Show it is linearly independent

1 4 -2 3
2 -2 1 1
3 1 2 11

I have the solution, but I don't understand.

Solution:

a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)

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why the above equation can change to the below equations??????

a + 2b + 3c = 0
4a - 2b + c = 0
-2a + b + 2 c = 0
3a + b + 11c = 0

a=b=c=0, so it is linearly independent.

Confused

2. Originally Posted by orange890
Linearly independent (MAtrix)

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The rows of the matrix:

1 4 -2 3
2 -2 1 1
3 1 2 11

I have the solution, but I don't understand.
You want to show that the rows of this matrix are linearly independent?

If they are linearly independent the only solution of:

a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)
is the trivial solution a=b=c=0.

a + 2b + 3c = 0
4a - 2b + c = 0
-2a + b + 2 c = 0
3a + b + 11c = 0
This is just your previous vector equation written as four scalar equations,
which you should proceed to solve to show that you must have: a=b=c=0,
and so you can conclude that the rows are linearly independent.

RonL

3. a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)
[COLOR=DarkOrange]
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How does it form??????

a + 2b + 3c = 0
4a - 2b + c = 0
-2a + b + 2 c = 0
3a + b + 11c = 0

4. Originally Posted by orange890
a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)
[COLOR=DarkOrange]
Look at the first component of the expressions on the Left and Right hand sides:

a + 2 b +3 c = 0

Now the seecond component:

4a-2b+c=0

Third component:

-2a+b+2c=0

Fourth component:

3a+b+11c=0

RonL

5. hi RonL,

yeah really thanks for your helping.